Question Number 3832 by Yozzii last updated on 21/Dec/15
$${Show}\:{that},\:\forall{a},{b}\in\mathbb{R}^{+} , \\ $$$$\:\left(\frac{{a}}{{b}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{{b}}{{a}}\right)^{\mathrm{1}/\mathrm{3}} \leqslant\left\{\mathrm{2}\left({a}+{b}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\right\}^{\mathrm{1}/\mathrm{3}} . \\ $$$$ \\ $$
Commented by RasheedSindhi last updated on 22/Dec/15
![((a/b))^(1/3) +((b/a))^(1/3) ≤{2(a+b)((1/a)+(1/b))}^(1/3) RHS:{2(a+b)((1/a)+(1/b))}^(1/3) ={2(a+b)(((a+b)/(ab)))}^(1/3) ={((2(a+b)^2 )/(ab))}^(1/3) =((2^(1/3) (a+b)^(2/3) )/((ab)^(1/3) )) LHS:((a/b))^(1/3) +((b/a))^(1/3) =(a^(1/3) /b^(1/3) )+(b^(1/3) /a^(1/3) )=((a^(2/3) +b^(2/3) )/((ab)^(1/3) )) LHS≤RHS ((a^(2/3) +b^(2/3) )/((ab)^(1/3) ))≤((2^(1/3) (a+b)^(2/3) )/((ab)^(1/3) )) ⇒a^(2/3) +b^(2/3) ≤2^(1/3) (a+b)^(2/3) [∵ a,b∈R^+ ] ⇒a^(2/3) +b^(2/3) ≤((√2))^(2/3) (a+b)^(2/3) ⇒a^(2/3) +b^(2/3) ≤((√2)a+(√2)b)^(2/3) Case(i) a=b a^(2/3) +b^(2/3) ≤((√2)a+(√2)b)^(2/3) ⇒2a^(2/3) ≤(2(√2)a)^(2/3) ⋮ Case(ii) a>b a^(2/3) +b^(2/3) ≤((√2)a+(√2)b)^(2/3) ⇒a^(2/3) (1+((b/a))^(2/3) )≤((√2)a)^(2/3) {1+(b/a)}^(2/3) ⇒a^(2/3) (1+((b/a))^(2/3) )≤2^(1/3) a^(2/3) {1+(b/a)}^(2/3) Dividing by a^(2/3) (>0) ⇒(1+((b/a))^(2/3) )≤2^(1/3) {1+(b/a)}^(2/3) Using binomial theorm (1+x)^n =1+nx+((n(n−1))/(2!))x^2 +... where ∣x∣<1 (1+(b/a))^(2/3) =1+((2/3))((b/a))^(2/3) +... Continue](https://www.tinkutara.com/question/Q3835.png)
$$\left(\frac{{a}}{{b}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{{b}}{{a}}\right)^{\mathrm{1}/\mathrm{3}} \leqslant\left\{\mathrm{2}\left({a}+{b}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\right\}^{\mathrm{1}/\mathrm{3}} \\ $$$${RHS}:\left\{\mathrm{2}\left({a}+{b}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$=\left\{\mathrm{2}\left({a}+{b}\right)\left(\frac{{a}+{b}}{{ab}}\right)\right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$=\left\{\frac{\mathrm{2}\left({a}+{b}\right)^{\mathrm{2}} }{{ab}}\right\}^{\mathrm{1}/\mathrm{3}} =\frac{\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left({a}+{b}\right)^{\mathrm{2}/\mathrm{3}} }{\left({ab}\right)^{\mathrm{1}/\mathrm{3}} } \\ $$$${LHS}:\left(\frac{{a}}{{b}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{{b}}{{a}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$=\frac{{a}^{\mathrm{1}/\mathrm{3}} }{{b}^{\mathrm{1}/\mathrm{3}} }+\frac{{b}^{\mathrm{1}/\mathrm{3}} }{{a}^{\mathrm{1}/\mathrm{3}} }=\frac{{a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} }{\left({ab}\right)^{\mathrm{1}/\mathrm{3}} } \\ $$$${LHS}\leqslant{RHS} \\ $$$$\frac{{a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} }{\left({ab}\right)^{\mathrm{1}/\mathrm{3}} }\leqslant\frac{\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left({a}+{b}\right)^{\mathrm{2}/\mathrm{3}} }{\left({ab}\right)^{\mathrm{1}/\mathrm{3}} } \\ $$$$\Rightarrow{a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} \leqslant\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left({a}+{b}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\because\:{a},{b}\in\mathbb{R}^{+} \right] \\ $$$$\Rightarrow{a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} \leqslant\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}/\mathrm{3}} \left({a}+{b}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\Rightarrow{a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} \leqslant\left(\sqrt{\mathrm{2}}{a}+\sqrt{\mathrm{2}}{b}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$${Case}\left({i}\right)\:{a}={b} \\ $$$${a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} \leqslant\left(\sqrt{\mathrm{2}}{a}+\sqrt{\mathrm{2}}{b}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \leqslant\left(\mathrm{2}\sqrt{\mathrm{2}}{a}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\vdots \\ $$$${Case}\left({ii}\right)\:{a}>{b} \\ $$$${a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} \leqslant\left(\sqrt{\mathrm{2}}{a}+\sqrt{\mathrm{2}}{b}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\Rightarrow{a}^{\mathrm{2}/\mathrm{3}} \left(\mathrm{1}+\left(\frac{{b}}{{a}}\right)^{\mathrm{2}/\mathrm{3}} \right)\leqslant\left(\sqrt{\mathrm{2}}{a}\right)^{\mathrm{2}/\mathrm{3}} \left\{\mathrm{1}+\frac{{b}}{{a}}\right\}^{\mathrm{2}/\mathrm{3}} \\ $$$$\Rightarrow{a}^{\mathrm{2}/\mathrm{3}} \left(\mathrm{1}+\left(\frac{{b}}{{a}}\right)^{\mathrm{2}/\mathrm{3}} \right)\leqslant\mathrm{2}^{\mathrm{1}/\mathrm{3}} {a}^{\mathrm{2}/\mathrm{3}} \left\{\mathrm{1}+\frac{{b}}{{a}}\right\}^{\mathrm{2}/\mathrm{3}} \\ $$$${Dividing}\:{by}\:{a}^{\mathrm{2}/\mathrm{3}} \left(>\mathrm{0}\right) \\ $$$$\Rightarrow\left(\mathrm{1}+\left(\frac{{b}}{{a}}\right)^{\mathrm{2}/\mathrm{3}} \right)\leqslant\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left\{\mathrm{1}+\frac{{b}}{{a}}\right\}^{\mathrm{2}/\mathrm{3}} \\ $$$${Using}\:{binomial}\:{theorm}\: \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}} =\mathrm{1}+{nx}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +… \\ $$$${where}\:\mid{x}\mid<\mathrm{1} \\ $$$$\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}/\mathrm{3}} =\mathrm{1}+\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\left(\frac{{b}}{{a}}\right)^{\mathrm{2}/\mathrm{3}} +… \\ $$$${Continue} \\ $$
Commented by Yozzii last updated on 22/Dec/15
$${There}'{s}\:{an}\:{error}\:{in}\:{the}\:{third}\:{line}\: \\ $$$${of}\:{your}\:{attempt}.\: \\ $$$$'…=\left\{\frac{\mathrm{2}\left({a}+{b}\right)^{\mathrm{2}} }{{ab}}\right\}^{\mathrm{1}/\mathrm{3}} =\frac{\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left({a}+{b}\right)^{\mathrm{2}/\mathrm{3}} }{\left({ab}\right)^{\mathrm{1}/\mathrm{3}} }…' \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 22/Dec/15
$$\mathcal{T}{han}\Bbbk\mathcal{S}!\:{I}\:{am}\:{going}\:{to}\:{correct}\:{it}. \\ $$
Answered by Rasheed Soomro last updated on 23/Dec/15
![Show that, ∀a,b∈R^+ , ((a/b))^(1/3) +((b/a))^(1/3) ≤{2(a+b)((1/a)+(1/b))}^(1/3) ..................(I) −−−−−−−−−−−−−−− A≤B ⇔A^3 ≤B^3 , ∀A,B>0 ((a/b))^(1/3) +((b/a))^(1/3) ≤{2(a+b)((1/a)+(1/b))}^(1/3) ⇒ [((a/b))^(1/3) +((b/a))^(1/3) ]^3 ≤[{2(a+b)((1/a)+(1/b))}^(1/3) ]^3 ⇒(a/b)+3((a/b))^(1/3) ((b/a))^(1/3) {((a/b))^(1/3) +((b/a))^(1/3) }+(b/a) ≤2(a+b)(((a+b)/(ab))) ⇒(a/b)+3(((a/b))^(1/3) +((b/a))^(1/3) )+(b/a)≤((2(a+b)^2 )/(ab)) ((a^2 +b^2 )/(ab))+3(((a^(2/3) +b^(2/3) )/(a^(1/3) b^(1/3) )))≤2(((a^2 +b^2 )/(ab)))+((4ab)/(ab)) 3(((a^(2/3) +b^(2/3) )/(a^(1/3) b^(1/3) )))≤((a^2 +b^2 )/(ab))+4 3(((a^(2/3) +b^(2/3) )/(a^(1/3) b^(1/3) )))ab≤a^2 +b^2 +4ab [Multiplying by ab>0] 3(a^(2/3) +b^(2/3) )a^(2/3) b^(2/3) ≤a^2 +b^2 +4ab Continue](https://www.tinkutara.com/question/Q3879.png)
$${Show}\:{that},\:\forall{a},{b}\in\mathbb{R}^{+} , \\ $$$$\:\left(\frac{{a}}{{b}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{{b}}{{a}}\right)^{\mathrm{1}/\mathrm{3}} \leqslant\left\{\mathrm{2}\left({a}+{b}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\right\}^{\mathrm{1}/\mathrm{3}} ………………\left(\boldsymbol{\mathrm{I}}\right) \\ $$$$−−−−−−−−−−−−−−− \\ $$$${A}\leqslant{B}\:\Leftrightarrow{A}^{\mathrm{3}} \leqslant{B}^{\mathrm{3}} \:\:,\:\:\:\forall{A},{B}>\mathrm{0} \\ $$$$\left(\frac{{a}}{{b}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{{b}}{{a}}\right)^{\mathrm{1}/\mathrm{3}} \leqslant\left\{\mathrm{2}\left({a}+{b}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\:\left[\left(\frac{{a}}{{b}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{{b}}{{a}}\right)^{\mathrm{1}/\mathrm{3}} \right]^{\mathrm{3}} \leqslant\left[\left\{\mathrm{2}\left({a}+{b}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\right\}^{\mathrm{1}/\mathrm{3}} \right]^{\mathrm{3}} \\ $$$$\Rightarrow\frac{{a}}{{b}}+\mathrm{3}\left(\frac{{a}}{{b}}\right)^{\mathrm{1}/\mathrm{3}} \left(\frac{{b}}{{a}}\right)^{\mathrm{1}/\mathrm{3}} \left\{\left(\frac{{a}}{{b}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{{b}}{{a}}\right)^{\mathrm{1}/\mathrm{3}} \right\}+\frac{{b}}{{a}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leqslant\mathrm{2}\left({a}+{b}\right)\left(\frac{{a}+{b}}{{ab}}\right) \\ $$$$\Rightarrow\frac{{a}}{{b}}+\mathrm{3}\left(\left(\frac{{a}}{{b}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{{b}}{{a}}\right)^{\mathrm{1}/\mathrm{3}} \right)+\frac{{b}}{{a}}\leqslant\frac{\mathrm{2}\left({a}+{b}\right)^{\mathrm{2}} }{{ab}} \\ $$$$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}}+\mathrm{3}\left(\frac{{a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} }{{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} }\right)\leqslant\mathrm{2}\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}}\right)+\frac{\mathrm{4}{ab}}{{ab}} \\ $$$$\mathrm{3}\left(\frac{{a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} }{{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} }\right)\leqslant\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}}+\mathrm{4} \\ $$$$\mathrm{3}\left(\frac{{a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} }{{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} }\right){ab}\leqslant{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{4}{ab}\:\:\left[{Multiplying}\:{by}\:{ab}>\mathrm{0}\right] \\ $$$$\mathrm{3}\left({a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} \right){a}^{\mathrm{2}/\mathrm{3}} {b}^{\mathrm{2}/\mathrm{3}} \leqslant{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{4}{ab} \\ $$$$ \\ $$$${Continue} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by Yozzii last updated on 24/Dec/15
$${Thanks}\:{for}\:{trying}\:{this}\:{question}.\: \\ $$$${It}\:{was}\:{taken}\:{from}\:{a}\:{national}\:{math} \\ $$$${olympiad}\:{exam}\:\left({Czech}\:{and}\:{Slovak}\right. \\ $$$$\left.{Republic}\right)\:{in}\:{the}\:{year}\:\mathrm{2000}.\: \\ $$$${I}\:{haven}'{t}\:{gotten}\:{anywhere}\:{near}\:{starting} \\ $$$${to}\:{solve}\:{the}\:{problem}.\: \\ $$
Commented by Rasheed Soomro last updated on 24/Dec/15
$$\mathcal{M}{y}\:\mathcal{E}{nglish}\:{is}\:{poor}.{So}\:\mathcal{I} \\ $$$$\mathcal{C}{ouldn}'{t}\:{understand}\:{sense}\:{of}\:{last} \\ $$$${two}\:\:{lines}.{Pl}\:{explain}. \\ $$
Commented by Yozzii last updated on 24/Dec/15
$${The}\:{problem}\:{is}\:{so}\:{difficult}\:{for}\:{me} \\ $$$${that},\:{it}\:{is}\:{taking}\:{a}\:{lot}\:{of}\:{time}\:{for}\:{me} \\ $$$${to}\:{solve}\:{it}.\: \\ $$
Commented by Rasheed Soomro last updated on 24/Dec/15
$$\mathbb{T}\mathrm{han}\Bbbk\mathbb{S}! \\ $$
Answered by Rasheed Soomro last updated on 23/Dec/15
![⇒1+((b/a))^(2/3) ≤2^(1/3) (1+(b/a))^(2/3) [∵a^(2/3) >0,dividing by it] ((a/b))^(1/3) +((b/a))^(1/3) ≤{2(a+b)((1/a)+(1/b))}^(1/3) (a^(1/3) /b^(1/3) )+(b^(1/3) /a^(1/3) )≤{2(a+b)(((a+b)/(ab)))}^(1/3) ((a^(2/3) +b^(2/3) )/((ab)^(1/3) ))≤{((2(a+b)^2 )/(ab))}^(1/3) ((a^(2/3) +b^(2/3) )/((ab)^(1/3) ))≤((2^(1/3) (a+b)^(2/3) )/((ab)^(1/3) )) a^(2/3) +b^(2/3) ≤2^(1/3) (a+b)^(2/3) (i) [∵(ab)^(1/3) >0 ,so by multiplying] Case(I) a=b (i) will be 2a^(2/3) ≤2^(1/3) (2a)^(2/3) 2a^(2/3) ≤2^(1/3) 2^(2/3) a^(2/3) 2a^(2/3) ≤2a^(2/3) ∴ ≤ holds Case(II) a>b (i) will be a^(2/3) (1+((b/a))^(2/3) )≤2^(1/3) a^(2/3) (1+(b/a))^(2/3) ⇒1+((b/a))^(2/3) ≤2^(1/3) (1+(b/a))^(2/3) [∵a^(2/3) >0,dividing by it] ⇒1+((b^2 /a^2 ))^(1/3) ≤2^(1/3) {(1+(b/a))^2 }^(1/3) ⇒1+((b^2 /a^2 ))^(1/3) ≤2^(1/3) (1+((2b)/a)+(b^2 /a^2 ))^(1/3) Let (b/a)=x then for x<1 1+x^(2/3) ≤2^(1/3) (1+2x+x^2 )^(1/3) Or 1+x^(2/3) ≤2^(1/3) (1+x)^(2/3) It couldn′t be solved by me.](https://www.tinkutara.com/question/Q3876.png)
$$\Rightarrow\mathrm{1}+\left(\frac{{b}}{{a}}\right)^{\mathrm{2}/\mathrm{3}} \leqslant\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}/\mathrm{3}} \:\:\:\:\left[\because{a}^{\mathrm{2}/\mathrm{3}} >\mathrm{0},{dividing}\:{by}\:{it}\right] \\ $$$$\:\left(\frac{{a}}{{b}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{{b}}{{a}}\right)^{\mathrm{1}/\mathrm{3}} \leqslant\left\{\mathrm{2}\left({a}+{b}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$\frac{{a}^{\mathrm{1}/\mathrm{3}} }{{b}^{\mathrm{1}/\mathrm{3}} }+\frac{{b}^{\mathrm{1}/\mathrm{3}} }{{a}^{\mathrm{1}/\mathrm{3}} }\leqslant\left\{\mathrm{2}\left({a}+{b}\right)\left(\frac{{a}+{b}}{{ab}}\right)\right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$\frac{{a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} }{\left({ab}\right)^{\mathrm{1}/\mathrm{3}} }\leqslant\left\{\frac{\mathrm{2}\left({a}+{b}\right)^{\mathrm{2}} }{{ab}}\right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$\frac{{a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} }{\left({ab}\right)^{\mathrm{1}/\mathrm{3}} }\leqslant\frac{\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left({a}+{b}\right)^{\mathrm{2}/\mathrm{3}} }{\left({ab}\right)^{\mathrm{1}/\mathrm{3}} } \\ $$$${a}^{\mathrm{2}/\mathrm{3}} +{b}^{\mathrm{2}/\mathrm{3}} \:\leqslant\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left({a}+{b}\right)^{\mathrm{2}/\mathrm{3}} \left({i}\right)\:\:\:\left[\because\left({ab}\right)^{\mathrm{1}/\mathrm{3}} >\mathrm{0}\:,{so}\:{by}\:{multiplying}\right] \\ $$$${Case}\left({I}\right)\:{a}={b} \\ $$$$\left({i}\right)\:{will}\:{be}\: \\ $$$$\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \leqslant\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{2}{a}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \leqslant\mathrm{2}^{\mathrm{1}/\mathrm{3}} \mathrm{2}^{\mathrm{2}/\mathrm{3}} {a}^{\mathrm{2}/\mathrm{3}} \\ $$$$\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \leqslant\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \\ $$$$\therefore\:\:\leqslant\:{holds}\: \\ $$$${Case}\left({II}\right)\:{a}>{b} \\ $$$$\left({i}\right)\:\:{will}\:{be} \\ $$$${a}^{\mathrm{2}/\mathrm{3}} \left(\mathrm{1}+\left(\frac{{b}}{{a}}\right)^{\mathrm{2}/\mathrm{3}} \right)\leqslant\mathrm{2}^{\mathrm{1}/\mathrm{3}} {a}^{\mathrm{2}/\mathrm{3}} \left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\Rightarrow\mathrm{1}+\left(\frac{{b}}{{a}}\right)^{\mathrm{2}/\mathrm{3}} \leqslant\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}/\mathrm{3}} \:\:\:\:\left[\because{a}^{\mathrm{2}/\mathrm{3}} >\mathrm{0},{dividing}\:{by}\:{it}\right] \\ $$$$\Rightarrow\mathrm{1}+\left(\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)^{\mathrm{1}/\mathrm{3}} \leqslant\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left\{\left(\mathrm{1}+\frac{{b}}{{a}}\right)^{\mathrm{2}} \right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$\Rightarrow\mathrm{1}+\left(\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)^{\mathrm{1}/\mathrm{3}} \leqslant\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{1}+\frac{\mathrm{2}{b}}{{a}}+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${Let}\:\frac{{b}}{{a}}={x}\:{then}\:{for}\:{x}<\mathrm{1} \\ $$$$\mathrm{1}+{x}^{\mathrm{2}/\mathrm{3}} \leqslant\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{3}} \\ $$$${Or}\:\mathrm{1}+{x}^{\mathrm{2}/\mathrm{3}} \leqslant\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{1}+{x}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\mathcal{I}{t}\:{couldn}'{t}\:{be}\:{solved}\:{by}\:{me}. \\ $$
Commented by RasheedSindhi last updated on 25/Dec/15
$${In}\:\mathrm{4}^{{th}} \:{line}\:{where}\:{does}\:\frac{\mathrm{2}}{\mathrm{3}}\:{disappear}? \\ $$
Commented by prakash jain last updated on 25/Dec/15
$$\mathrm{1}+{x}^{\mathrm{2}/\mathrm{3}} \leqslant\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{1}+{x}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$${f}\left({x}\right)=\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{1}+{x}\right)^{\mathrm{2}/\mathrm{3}} −\mathrm{1}−{x}^{\mathrm{2}/\mathrm{3}} \\ $$$${f}\:'\left({x}\right)=\mathrm{2}^{\mathrm{1}/\mathrm{3}} \frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}+{x}\right)^{−\mathrm{1}/\mathrm{3}} −\frac{\mathrm{2}}{\mathrm{3}}{x}^{−\mathrm{1}/\mathrm{3}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{1}+{x}\right)^{−\mathrm{1}/\mathrm{3}} −{x}^{−\mathrm{1}/\mathrm{3}} \right) \\ $$$$\frac{\mathrm{2}}{\mathrm{1}+{x}}−\frac{\mathrm{1}}{{x}}=\frac{\mathrm{2}{x}−\mathrm{1}−{x}}{{x}\left(\mathrm{1}+{x}\right)}<\mathrm{0}\:\mathrm{for}\:\mathrm{0}<{x}<\mathrm{1} \\ $$$$\frac{\mathrm{2}}{\mathrm{1}+{x}}<\frac{\mathrm{1}}{{x}}\Rightarrow\mathrm{2}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{1}+{x}\right)^{−\mathrm{1}/\mathrm{3}} <{x}^{−\mathrm{1}/\mathrm{3}} \\ $$$${f}'\left({x}\right)\leqslant\mathrm{0}\:\mathrm{0}<{x}<\mathrm{1} \\ $$$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{decreasing}. \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{2}^{\mathrm{1}/\mathrm{3}} −\mathrm{1}\geqslant\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{Since}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{strictly}\:\mathrm{decreasing}\:\mathrm{for}\:\mathrm{0}<{x}<\mathrm{1} \\ $$$$\mathrm{and}\:{f}\left(\mathrm{0}\right)>\mathrm{0}\:\mathrm{and}\:{f}\left(\mathrm{1}\right)=\mathrm{0}{f}\left({x}\right)>\mathrm{0}\:{for}\:\mathrm{0}<{x}<\mathrm{1}. \\ $$$${b}>{a}\:{need}\:{not}\:{be}\:{checked}\:{since}\:{inequality} \\ $$$${is}\:{symmetric}\:{in}\:{a}\:{and}\:{b}. \\ $$$${x}={b}=\mathrm{0}\:\mathrm{LHS}\:\mathrm{and}\:\mathrm{RHS}\:\mathrm{are}\:\mathrm{undefined}. \\ $$
Commented by prakash jain last updated on 25/Dec/15
$$\mathrm{We}\:\mathrm{are}\:\mathrm{only}\:\mathrm{interested}\:\mathrm{in}\:\mathrm{sign}. \\ $$
Commented by Rasheed Soomro last updated on 25/Dec/15
$$\mathbb{T}\mathrm{h}\alpha{n}\Bbbk\mathfrak{S}! \\ $$