# Prove-that-among-all-triangle-of-equal-perimeter-equilateral-triangle-has-the-largest-area-

Question Number 3324 by prakash jain last updated on 10/Dec/15
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{among}\:\mathrm{all}\:\mathrm{triangle}\:\mathrm{of}\:\mathrm{equal} \\$$$$\mathrm{perimeter},\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{has}\:\mathrm{the} \\$$$$\mathrm{largest}\:\mathrm{area}. \\$$
Commented by 123456 last updated on 10/Dec/15
$$\mathrm{2}{s}\left({x},{y},{z}\right)={x}+{y}+{z} \\$$$$\mathrm{A}\left({x},{y},{z}\right)=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)} \\$$
Commented by Rasheed Soomro last updated on 11/Dec/15
$${Let}\:{x},{x}+{l}\:\:{and}\:\:{x}+{l}+{m}\:\:{are}\:{measures}\:{of}\:\:{three}\:{sides} \\$$$${where}\:{l},{m}\geqslant\mathrm{0} \\$$$$\:\:\:\:\:\:{Perimeter}=\mathrm{3}{x}+\mathrm{2}{l}+{m} \\$$$$\:\:\:\:\:\:\:\:\:{S}=\frac{\mathrm{3}{x}+\mathrm{2}{l}+{m}}{\mathrm{2}} \\$$$$\mathcal{I}{n}\:\mathcal{E}{quilateral}\:{triangle}\:{S}=\frac{\mathrm{3}{x}}{\mathrm{2}}\:\:{for}\:{l},{m}=\mathrm{0} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{A}_{\bigtriangleup} =\sqrt{\frac{\mathrm{3}{x}}{\mathrm{2}}\left(\frac{\mathrm{3}{x}}{\mathrm{2}}−{x}\right)^{\mathrm{3}} }=\sqrt{\frac{\mathrm{3}{x}}{\mathrm{2}}×\frac{{x}^{\mathrm{3}} }{\mathrm{2}^{\mathrm{3}} }} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\frac{\mathrm{3}{x}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{4}} }}=\frac{\sqrt{\mathrm{3}}\:{x}^{\mathrm{2}} }{\mathrm{4}} \\$$$${As}\:{S}\:{is}\:{constant}\:\left({because}\:{perimeter}\:{is}\:{constant}\right. \\$$$${Hence}\:{for}\:{any}\:{triangle}\:{S} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:{S}=\frac{\mathrm{3}{x}}{\mathrm{2}} \\$$$${Area}\:{of}\:{any}\:{triangle}\: \\$$$$\:\:{A}=\sqrt{\frac{\mathrm{3}{x}}{\mathrm{2}}\left(\frac{\mathrm{3}{x}}{\mathrm{2}}−{x}\right)\left(\frac{\mathrm{3}{x}}{\mathrm{2}}−{x}−{l}\right)\left(\frac{\mathrm{3}{x}}{\mathrm{2}}−{x}−{l}−{m}\right)} \\$$$$\:\:\:\:\:\:\:=\sqrt{\frac{\mathrm{3}{x}}{\mathrm{2}}×\frac{{x}}{\mathrm{2}}×\frac{{x}−\mathrm{2}{l}}{\mathrm{2}}×\frac{{x}−\mathrm{2}{l}−\mathrm{2}{m}}{\mathrm{2}}} \\$$$$\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}{x}^{\mathrm{2}} \left({x}−\mathrm{2}{l}\right)\left({x}−\mathrm{2}{l}−\mathrm{2}{m}\right)}}{\mathrm{4}} \\$$$${We}\:{have}\:{to}\:{prove}\: \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:{A}_{\bigtriangleup} \geqslant{A} \\$$$$\frac{\sqrt{\mathrm{3}}\:{x}^{\mathrm{2}} }{\mathrm{4}}\geqslant\frac{\sqrt{\mathrm{3}{x}^{\mathrm{2}} \left({x}−\mathrm{2}{l}\right)\left({x}−\mathrm{2}{l}−\mathrm{2}{m}\right)}}{\mathrm{4}}\:\:{for}\:{l},{m}\geqslant\mathrm{0} \\$$$${C}^{\mathcal{ONTINUE}} \\$$
Commented by prakash jain last updated on 11/Dec/15
$$\mathrm{From}\:\mathrm{your}\:\mathrm{derivation} \\$$$$\mathrm{In}\:\mathrm{your}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{area} \\$$$${a}={x},\:{b}={x}−{l},\:{c}={x}−{l}−{m} \\$$$${how}\:{s}=\frac{\mathrm{3}{x}}{\mathrm{2}} \\$$
Commented by Rasheed Soomro last updated on 12/Dec/15
$${For}\:{equilateral}\:{triangle}\:{of}\:{side}\:{x}\:\:{S}=\frac{\mathrm{3}{x}}{\mathrm{2}} \\$$$${and}\:{I}\:{have}\:{used}\:{this}\:{also}\:\:{in}\:{case}\:{of}\:{common} \\$$$${triangle},{because}\:{the}\:{perimeter}\:{is}\:{constant}. \\$$$$\mathcal{B}{ut}\:{now}\:{I}\:{have}\:{changed}\:{my}\:{mind}.\:{I}\:{am}\:{working} \\$$$${on}\:{different}\:{way}.\:{Pl}\:{see}\:{my}\:{answer}\:{below}. \\$$$${I}\:{am}\:{always}\:\:\boldsymbol{\mathrm{thankful}}\:{to}\:{you}\:{for}\:{guidance}. \\$$
Answered by Rasheed Soomro last updated on 13/Dec/15
$$\mathcal{A}{rea}\:\:{of}\:\:\mathcal{C}{ommon}\:\mathcal{T}{riangle} \\$$$${Let}\:{the}\:{measures}\:{of}\:{sides}\:{of}\:\:{a}\:{common}\:{triangle}\:{are} \\$$$${x},{y},{z}\: \\$$$$\mathcal{A}{rea}\left({A}\right)\:{will}\:{be} \\$$$$\:\:\:\:\:\:{A}=\sqrt{{S}\left({S}−{x}\right)\left({S}−{y}\right)\left({S}−{x}\right)}\:\:\:\:;\:\:{S}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+{y}+{z}\right) \\$$$$\:\:\:\:=\sqrt{\frac{{x}+{y}+{z}}{\mathrm{2}}\left(\frac{{x}+{y}+{z}}{\mathrm{2}}−{x}\right)\left(\frac{{x}+{y}+{z}}{\mathrm{2}}−{y}\right)\left(\frac{{x}+{y}+{z}}{\mathrm{2}}−{z}\right)} \\$$$$\:\:\:\:=\sqrt{\left(\frac{{x}+{y}+{z}}{\mathrm{2}}\right)\left(\frac{{x}+{y}+{z}−\mathrm{2}{x}}{\mathrm{2}}\right)\left(\frac{{x}+{y}+{z}−\mathrm{2}{y}}{\mathrm{2}}\right)\left(\frac{{x}+{y}+{z}−\mathrm{2}{z}}{\mathrm{2}}\right)} \\$$$$\:\:\:\:=\sqrt{\left(\frac{{x}+{y}+{z}}{\mathrm{2}}\right)\left(\frac{−{x}+{y}+{z}}{\mathrm{2}}\right)\left(\frac{{x}−{y}+{z}}{\mathrm{2}}\right)\left(\frac{{x}+{y}−{z}}{\mathrm{2}}\right)} \\$$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\left({x}+{y}+{z}\right)\left(−{x}+{y}+{z}\right)\left({x}−{y}+{z}\right)\left({x}+{y}−{z}\right)} \\$$$$\mathcal{A}{rea}\:\:{of}\:\:\mathcal{E}{quilateral}\:\mathcal{T}{riangle} \\$$$$\therefore\:{Every}\:{side}\:\:{of}\:\:{of}\:{an}\:{equilateral}\:{triangle}\:{having}\: \\$$$${same}\:{perimeter}\:\left({x}+{y}+{z}\right)\:{is}\:\:\frac{{x}+{y}+{z}}{\mathrm{3}} \\$$$$\:\therefore\:\:\:{A}_{\bigtriangleup} =\sqrt{{S}\left({S}−\:\frac{{x}+{y}+{z}}{\mathrm{3}}\right)^{\mathrm{3}} } \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\frac{{x}+{y}+{z}}{\mathrm{2}}×\left(\frac{{x}+{y}+{z}}{\mathrm{2}}−\frac{{x}+{y}+{z}}{\mathrm{3}}\right)^{\mathrm{3}} } \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\frac{{x}+{y}+{z}}{\mathrm{2}}×\left(\frac{\mathrm{3}{x}+\mathrm{3}{y}+\mathrm{3}{z}−\mathrm{2}{x}−\mathrm{2}{y}−\mathrm{2}{z}}{\mathrm{6}}\right)^{\mathrm{3}} } \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\frac{{x}+{y}+{z}}{\mathrm{2}}×\left(\frac{{x}+{y}+{z}}{\mathrm{6}}\right)^{\mathrm{3}} } \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\frac{\left({x}+{y}+{z}\right)^{\mathrm{4}} }{\mathrm{2}^{\mathrm{4}} ×\mathrm{3}^{\mathrm{2}} ×\mathrm{3}}}=\frac{\left({x}+{y}+{z}\right)^{\mathrm{2}} }{\mathrm{12}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\$$$$\:\:\:\:\:\:\:\:{A}_{\bigtriangleup} =\frac{\left({x}+{y}+{z}\right)^{\mathrm{2}} }{\mathrm{12}\sqrt{\mathrm{3}}} \\$$$$−−−−−<><><><>−−−−− \\$$$${Now}\:{we}\:{have}\:{to}\:{prove}: \\$$$${A}_{\bigtriangleup} \geqslant{A} \\$$$$\:\:\frac{\left({x}+{y}+{z}\right)^{\mathrm{2}} }{\mathrm{12}\sqrt{\mathrm{3}}}\geqslant\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\left({x}+{y}+{z}\right)\left(−{x}+{y}+{z}\right)\left({x}−{y}+{z}\right)\left({x}+{y}−{z}\right)} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{for}\:{x},{y},{z}>\mathrm{0} \\$$$$\Rightarrow\frac{\left({x}+{y}+{z}\right)^{\mathrm{2}} }{\mathrm{3}\sqrt{\mathrm{3}}}\geqslant\sqrt{\left({x}+{y}+{z}\right)\left(−{x}+{y}+{z}\right)\left({x}−{y}+{z}\right)\left({x}+{y}−{z}\right)} \\$$$$\boldsymbol{\mathrm{Yozzi}}'{s}\:\boldsymbol{\mathrm{contribution}}: \\$$$${Yozzi}\:{has}\:{proved}\:{that} \\$$$$\Rightarrow\frac{\left({x}+{y}+{z}\right)^{\mathrm{2}} }{\mathrm{4}}\geqslant\sqrt{\left({x}+{y}+{z}\right)\left(−{x}+{y}+{z}\right)\left({x}−{y}+{z}\right)\left({x}+{y}−{z}\right)} \\$$$$\left({See}\:{answer}\:\:{by}\:{Yozzi}.\:{An}\:{approach}\:{I}\:{can}'{t}\:{even}\right. \\$$$$\left.{think}\:{of}!\right) \\$$$${Now}\:{since}\:\mathrm{4}<\mathrm{3}\sqrt{\mathrm{3}} \\$$$$\therefore\:\:\:\:\frac{\left({x}+{y}+{z}\right)^{\mathrm{2}} }{\mathrm{3}\sqrt{\mathrm{3}}}\geqslant\sqrt{\left({x}+{y}+{z}\right)\left(−{x}+{y}+{z}\right)\left({x}−{y}+{z}\right)\left({x}+{y}−{z}\right)} \\$$$$\therefore\:{A}_{\bigtriangleup} \geqslant{A} \\$$$${QED} \\$$
Commented by Rasheed Soomro last updated on 14/Dec/15
$$\frac{\left({x}+{y}+{z}\right)^{\mathrm{2}} }{\mathrm{4}}\geqslant\sqrt{\underset{{r}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}{a}_{{r}} } \\$$$${it}\:{cannot}\:{solely}\:{lead}\:{one}\:{to}\:{safely}\:{stating} \\$$$${that}\:\frac{\left({x}+{y}+{z}\right)^{\mathrm{2}} }{\mathrm{3}\sqrt{\mathrm{3}}}\geqslant\sqrt{\underset{{r}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}{a}_{{r}} }\:. \\$$$$\vdots \\$$$${Why}\:{transitive}\:{property}\:{of}\:{inequality}\:{doesn}'{t} \\$$$${help}\:? \\$$$${If}\:{we}\:{have}\:{to}\:{prove}\:{that}\:{A}>{B}\:\:{and}\:\:{we}\:{have} \\$$$${by}\:{some}\:{way}\:\frac{{A}}{\mathrm{2}}>{B} \\$$$$\frac{{A}}{\mathrm{2}}>{B}\:\Rightarrow\:{A}>{B}\:? \\$$$$\\$$
Commented by Yozzii last updated on 13/Dec/15
$${If}\:\mathrm{4}<\mathrm{3}\sqrt{\mathrm{3}}\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}>\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}\Rightarrow\frac{\left({x}+{y}+{z}\right)^{\mathrm{2}} }{\mathrm{4}}>\frac{\left({x}+{y}+{z}\right)^{\mathrm{2}} }{\mathrm{3}\sqrt{\mathrm{3}}} \\$$$${Now},\:{we}\:{really}\:{weren}'{t}\:{given}\:{that} \\$$$${A}_{\bigtriangleup} \geqslant{A}.\:{Even}\:{if}\:{we}\:{were}\:{to}\:{check}\: \\$$$${that}\:{this}\:{is}\:{the}\:{case},\:{we}\:{would}\:{have} \\$$$${to}\:{decompose}\:{the}\:{inequality}\:{to}\:{a} \\$$$${form}\:{that}\:{the}\:{information}\:{we}\:{have} \\$$$${about}\:{the}\:{situation}\:{is}\:{satisfies}.\:{The} \\$$$${AM}−{GM}\:{sadly}\:{didn}'{t}\:{help}\:{as}\:{I}\:{had} \\$$$${hoped}\:{it}\:{would}\:{because}\:{while}\:{it} \\$$$${shows}\:{that}\: \\$$$$\frac{\left({x}+{y}+{z}\right)^{\mathrm{2}} }{\mathrm{4}}\geqslant\sqrt{\underset{{r}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}{a}_{{r}} } \\$$$${it}\:{cannot}\:{solely}\:{lead}\:{one}\:{to}\:{safely}\:{stating} \\$$$${that}\:\frac{\left({x}+{y}+{z}\right)^{\mathrm{2}} }{\mathrm{3}\sqrt{\mathrm{3}}}\geqslant\sqrt{\underset{{r}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}{a}_{{r}} }\:. \\$$$$\\$$$${The}\:{transitive}\:{property}\:{of}\:{inequalities} \\$$$${wouldn}'{t}\:{have}\:{worked}\:{for}\:{us}\:{in}\:{this} \\$$$${approach}\::\left(\:.\right. \\$$$$\\$$$$\\$$$$\\$$
Commented by Yozzii last updated on 14/Dec/15
$${Say}\:{we}\:{have}\:{the}\:{following}\:\:{problem}. \\$$$$−−−−−−−−−−−−−−−−−−−−− \\$$$${We}\:{want}\:{to}\:{show}\:{p}>{q}.\:{We}\:{have}\:{that} \\$$$${r}>{q}\:{and}\:{r}={np}\:{where}\:{n},{p},{q}\in\mathbb{R}^{+} \\$$$$\therefore\:\:{np}>{q}\Rightarrow{p}>\frac{\mathrm{1}}{{n}}{q}.\:\:{q}=\sqrt{\underset{{r}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}{a}_{{r}} }. \\$$$$\\$$$${Let}\:{p}=\frac{{u}}{\mathrm{3}\sqrt{\mathrm{3}}}\:{and}\:{r}=\frac{{u}}{\mathrm{4}}\:,{u}=\left({x}+{y}+{z}\right)^{\mathrm{2}} . \\$$$${u}=\mathrm{3}{p}\sqrt{\mathrm{3}}\Rightarrow{r}=\frac{\mathrm{3}{p}\sqrt{\mathrm{3}}}{\mathrm{4}}\Rightarrow\:{n}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}. \\$$$$\therefore\:{p}>\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}}{q}\: \\$$$$\frac{\mathrm{16}}{\mathrm{27}}=\left(\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \Rightarrow\:\mathrm{0}<\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}}<\mathrm{1}. \\$$$$\therefore\:{The}\:{information}\:{only}\:{shows}\:{that} \\$$$${p}>\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}}{q}\:{and}\:{not}\:{p}>{q}\:{since}\:\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}}{q}<{q}. \\$$$${On}\:{a}\:{number}\:{line}\:{we}\:{have}\:{three}\:{cases} \\$$$${due}\:{to}\:{this}\:{result}: \\$$$$\left({i}\right)\:\:\:\overset{\rightarrow} {\mathrm{0}}……………\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}}{q}……..{p}….{q}…. \\$$$$\left({ii}\right)\:\overset{\rightarrow} {\mathrm{0}}……………\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}}{q}……..{q}….{p}….. \\$$$$\left({iii}\right)\overset{\rightarrow} {\mathrm{0}}…………..\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}}{q}…….{p}={q}…… \\$$$${Which}\:{is}\:{correct}? \\$$
Commented by Rasheed Soomro last updated on 15/Dec/15
$$\mathcal{T}^{\mathcal{H}^{\mathcal{A}} \mathcal{N}} \mathcal{K}_{\mathcal{S}{sss}} ! \\$$
Commented by Yozzii last updated on 14/Dec/15
$${Let}\:{r}>\mathrm{7}\:{and}\:{r}=\mathrm{1}.\mathrm{2}{p}. \\$$$$\therefore\mathrm{1}.\mathrm{2}{p}>\mathrm{7}\Rightarrow{p}>\frac{\mathrm{7}}{\mathrm{1}.\mathrm{2}}\:{but}\:{this} \\$$$${doesn}'{t}\:{show}\:{that}\:{p}>\mathrm{7}. \\$$$$\\$$$$\:{Similarly}, \\$$$$\frac{\left({z}+{y}+{x}\right)^{\mathrm{2}} }{\mathrm{4}}\geqslant\sqrt{\underset{{r}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}{a}_{{r}} } \\$$$$\Rightarrow\left({x}+{y}+{z}\right)^{\mathrm{2}} \geqslant\mathrm{4}\sqrt{\underset{{r}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}{a}_{{r}} } \\$$$$\Rightarrow\frac{\left({x}+{y}+{z}\right)^{\mathrm{2}} }{\mathrm{3}\sqrt{\mathrm{3}}}\geqslant\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}}\sqrt{\underset{{r}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}{a}_{{r}} }\ngeq\sqrt{\underset{{r}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}{a}_{{r}} }\:{since}\:\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}}\ngtr\mathrm{1} \\$$$$\\$$
Commented by Rasheed Soomro last updated on 15/Dec/15
$$\boldsymbol{\mathcal{T}}\mathcal{H}\alpha{n}\boldsymbol{\mathcal{N}}\Bbbk\mathcal{S}^{\mathcal{SS}{Sssss}!} \\$$
Answered by Yozzi last updated on 12/Dec/15
$${For}\:{a}\:{triangle}\:{with}\:{side}\:{lengths} \\$$$${x},{y}\:{and}\:{z},\:{the}\:{length}\:{of}\:{one}\:{side} \\$$$${cannot}\:{exceed}\:{the}\:{sum}\:{of}\:{the}\:{lengths} \\$$$${of}\:{the}\:{other}\:{two}\:{sides}.\:{Symbolically}, \\$$$${x}+{y}−{z}\geqslant\mathrm{0}\:\wedge\:{x}+{z}−{y}\geqslant\mathrm{0}\:\wedge\:{z}+{y}−{x}\geqslant\mathrm{0}. \\$$$$\\$$$$\mathrm{PROOF}:\:{Suppose}\:{for}\:{contradiction} \\$$$${that}\:{one}\:{side}\:{of}\:{length}\:\boldsymbol{{a}}\:{of}\:{a}\:{given} \\$$$${triangle}\:{is}\:{more}\:{than}\:{the}\:{sum}\:{of}\:{the} \\$$$${lengths}\:{of}\:{the}\:{other}\:{two}\:{sides},\:{denoted} \\$$$${by}\:\boldsymbol{{b}}\:{and}\:\boldsymbol{{c}}.\: \\$$$$\therefore\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{a}}>\boldsymbol{{b}}+\boldsymbol{{c}} \\$$$${with}\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}>\mathrm{0}. \\$$$$\Rightarrow\:\boldsymbol{{a}}^{\mathrm{2}} >\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{bc}} \\$$$$\Rightarrow\:\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} >\mathrm{2}\boldsymbol{{bc}}\:\:\:\:\:\:\left(\ast\right) \\$$$${By}\:{the}\:{cosine}\:{rule},\: \\$$$$\boldsymbol{{a}}^{\mathrm{2}} =\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{bc}}{cos}\theta \\$$$${where}\:\theta\:{is}\:{the}\:{angle}\:{subtended}\:{by}\:{the} \\$$$${meeting}\:{of}\:{the}\:{sides}\:\boldsymbol{{b}}\:{and}\:\boldsymbol{{c}},\:{opposite} \\$$$${to}\:\boldsymbol{{a}}. \\$$$$\Rightarrow\:\:−\mathrm{2}\boldsymbol{{bc}}{cos}\theta=\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \\$$$$\therefore\:\:−\mathrm{2}\boldsymbol{{bc}}{cos}\theta>\mathrm{2}\boldsymbol{{bc}}\Rightarrow{cos}\theta<−\mathrm{1} \\$$$$\Rightarrow\:\mid{cos}\theta\mid>\mathrm{1}.\:{But},\:\forall\theta\in\mathbb{R},\:\mid{cos}\theta\mid\leqslant\mathrm{1}. \\$$$${This}\:{is}\:{a}\:{contradiction},\:{so}\:{that}\: \\$$$${indeed}\:\boldsymbol{{a}}\leqslant\boldsymbol{{b}}+\boldsymbol{{c}}\:{in}\:{general}\:{and}\:{we} \\$$$${can}\:{show}\:{that}\:\boldsymbol{{b}}\leqslant\boldsymbol{{a}}+\boldsymbol{{c}}\:{and}\:\boldsymbol{{c}}\leqslant\boldsymbol{{a}}+\boldsymbol{{b}} \\$$$${likewise},\:{for}\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}>\mathrm{0}.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Box \\$$$${Hence},\:{all}\:{terms}\:{under}\:{the}\:{square} \\$$$${root}\:{of}\:{Mr}.\:{Rasheed}'{s}\:{rhs}\:{expression} \\$$$${are}\:{non}−{negative}.\: \\$$$$\\$$$$\\$$$${Since}\:{all}\:{terms}\:{are}\:{positive}\:{reals},\:{by}\:{AM}−{GM}, \\$$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\left\{{x}+{y}+{z}\right\}+\left\{{x}+{y}−{z}\right\}+\left\{{x}+{z}−{y}\right\}+\left\{{z}+{y}−{x}\right\}\right)\geqslant\left(\left({x}+{y}+{z}\right)\left({x}+{y}−{z}\right)\left({z}+{y}−{x}\right)\left({x}+{z}−{y}\right)\right)^{\mathrm{1}/\mathrm{4}} \\$$$${a}_{\mathrm{1}} ={x}+{y}+{z},\:\:{a}_{\mathrm{2}} ={x}+{y}−{z} \\$$$${a}_{\mathrm{3}} ={x}+{z}−{y},\:\:{a}_{\mathrm{4}} ={z}+{y}−{x} \\$$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}\left({x}+{y}+{z}\right)\right)\geqslant\left(\underset{{r}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}{a}_{{r}} \right)^{\mathrm{1}/\mathrm{4}} \\$$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\left({x}+{y}+{z}\right)^{\mathrm{2}} \geqslant\sqrt{\underset{{r}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}{a}_{{r}} } \\$$$$\\$$$$\\$$
Commented by Rasheed Soomro last updated on 12/Dec/15
$$\mathcal{I}\:\:{didn}'{t}\:{understand}\:{last}\:{two}\:{steps}. \\$$$$\mathcal{W}{here}\:{has}\:{a}_{{r}} \:{come}\:\:{from}? \\$$
Commented by Yozzi last updated on 12/Dec/15
$${I}'{ve}\:{added}\:{definitions}\:{for}\:{a}_{{r}} ,\:\mathrm{1}\leqslant{r}\leqslant\mathrm{4}. \\$$