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Question Number 221233 by Nicholas666 last updated on 28/May/25
if a,b,c,d,e,f > 0 and abcdef = 1 , then (1/( (√(1 + ad)))) + (1/( (√(1 + be)))) + (1/( (√(1 + cf)))) ≤ (3/( (√2))) Profosed by Craciun Georghe
$$ \\ $$$$\:\mathrm{if}\:{a},{b},{c},{d},{e},{f}\:>\:\mathrm{0}\:\mathrm{and}\:{abcdef}\:=\:\mathrm{1}\:, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{then} \\ $$$$\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:{ad}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:{be}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:{cf}}}\:\leqslant\:\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\: \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Profosed}\:\mathrm{by}\:\mathrm{Craciun}\:\mathrm{Georghe} \\ $$
Commented by Rasheed.Sindhi last updated on 28/May/25
a,b,c,d,e,f > 0 ⇒abcdef ≠ 0
$$\:\:\:\:\:\:\:\:{a},{b},{c},{d},{e},{f}\:>\:\mathrm{0}\:\Rightarrow{abcdef}\:\neq\:\mathrm{0} \\ $$
Commented by Nicholas666 last updated on 28/May/25
thank you for your correction Sir
$$\:\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{your}\:\mathrm{correction}\:\mathrm{Sir} \\ $$$$ \\ $$
Answered by vnm last updated on 31/May/25
let A=ad, B=be, C=cf A,B,C>0, ABC=1. Among the numbers A, B, C there are at least two such that their product is not greater than1. Let these numbers be A, B. let AB=q^2 , 0<q<1. A=qx, B=(q/x), C=(1/q^2 ), 0<x<1. let (1/( (√(1+A))))+(1/( (√(1+B))))=f(x)=(1/( (√(1+qx))))+(1/( (√(1+(q/x))))) Prove, that max_(0<x≤1) f(x)=f(1)=(2/( (√(1+q)))). (d/dx)f(x)=−(q/2)(1+qx)^(−(3/2)) +(q/(2x^2 ))∙(1+(q/x))^(−(3/2)) = (q/2)((1/x^2 )(1+(q/x))^(−(3/2)) −(1+qx)^(−(3/2)) ) we have to prove that (1/x^2 )(1+(q/x))^(−(3/2)) ≥_(0<x≤1) (1+qx)^(−(3/2)) x^2 (1+(q/x))^(3/2) ≤(1+qx)^(3/2) x^4 (1+(q/x))^3 ≤(1+qx)^3 x(x+q)^3 ≤(1+qx)^3 (1+qx)^3 −x(x+q)^3 =(1−x^2 )(1+x^2 +xq(3−q^2 ))≥0 : TRUE We have proved that (d/dx)f(x)≥0 ⇒ (1/( (√(1+qx))))+(1/( (√(1+(q/x))))) ≤_(0<x≤1) (2/( (√(1+q)))). (1/( (√(1+A))))+(1/( (√(1+B))))+(1/( (√(1+C))))≤(2/( (√(1+q))))+(1/( (√(1+(1/q^2 ))))) let g(q)=(2/( (√(1+q))))+(1/( (√(1+(1/q^2 ))))) prove that (d/dq)g(q)≥_(0<q≤1) 0. (d/dq)((2/( (√(1+q))))+(q/( (√(1+q^2 )))))= 2(−(1/2))(1+q)^(−(3/2)) +(1+q^2 )^(−(1/2)) +q(−(1/2))(1+q^2 )^(−(3/2)) 2q= (1+q^2 )^(−(1/2)) −q^2 (1+q^2 )^(−(3/2)) −(1+q)^(−(3/2)) = (((1+q^2 )−q^2 )/((1+q^2 )^(3/2) ))−(1/((1+q)^(3/2) ))=(1/((1+q^2 )^(3/2) ))−(1/((1+q)^(3/2) )) ≥_(0<q≤1) 0 ⇒ max_(0<q≤1) g(q)=g(1)=(2/( (√(1+1))))+(1/( (√(1+1))))=(3/( (√2))).
$$ \\ $$$$\mathrm{let}\:{A}={ad},\:{B}={be},\:{C}={cf} \\ $$$${A},{B},{C}>\mathrm{0},\:\:{ABC}=\mathrm{1}. \\ $$$$\mathrm{Among}\:\mathrm{the}\:\mathrm{numbers}\:\mathrm{A},\:\mathrm{B},\:\mathrm{C} \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{at}\:\mathrm{least}\:\mathrm{two}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{their}\:\mathrm{product}\:\mathrm{is}\:\mathrm{not}\:\mathrm{greater}\: \\ $$$$\mathrm{than1}.\:\mathrm{Let}\:\mathrm{these}\:\mathrm{numbers}\:\mathrm{be}\:\mathrm{A},\:\mathrm{B}. \\ $$$$\mathrm{let}\:{AB}={q}^{\mathrm{2}} ,\:\mathrm{0}<{q}<\mathrm{1}. \\ $$$${A}={qx},\:{B}=\frac{{q}}{{x}},\:{C}=\frac{\mathrm{1}}{{q}^{\mathrm{2}} },\:\mathrm{0}<{x}<\mathrm{1}. \\ $$$$\mathrm{let}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{A}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{B}}}={f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{qx}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{q}}{{x}}}} \\ $$$$\mathrm{Prove},\:\mathrm{that}\:\underset{\mathrm{0}<{x}\leqslant\mathrm{1}} {\mathrm{max}}{f}\left({x}\right)={f}\left(\mathrm{1}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}+{q}}}. \\ $$$$\frac{{d}}{{dx}}{f}\left({x}\right)=−\frac{{q}}{\mathrm{2}}\left(\mathrm{1}+{qx}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} +\frac{{q}}{\mathrm{2}{x}^{\mathrm{2}} }\centerdot\left(\mathrm{1}+\frac{{q}}{{x}}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} = \\ $$$$\frac{{q}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{q}}{{x}}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} −\left(\mathrm{1}+{qx}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\mathrm{1}+\frac{{q}}{{x}}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \:\:\underset{\mathrm{0}<{x}\leqslant\mathrm{1}} {\geqslant}\:\:\left(\mathrm{1}+{qx}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \: \\ $$$${x}^{\mathrm{2}} \left(\mathrm{1}+\frac{{q}}{{x}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \leqslant\left(\mathrm{1}+{qx}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${x}^{\mathrm{4}} \left(\mathrm{1}+\frac{{q}}{{x}}\right)^{\mathrm{3}} \leqslant\left(\mathrm{1}+{qx}\right)^{\mathrm{3}} \\ $$$${x}\left({x}+{q}\right)^{\mathrm{3}} \leqslant\left(\mathrm{1}+{qx}\right)^{\mathrm{3}} \\ $$$$\left(\mathrm{1}+{qx}\right)^{\mathrm{3}} −{x}\left({x}+{q}\right)^{\mathrm{3}} =\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} +{xq}\left(\mathrm{3}−{q}^{\mathrm{2}} \right)\right)\geqslant\mathrm{0}\::\:{TRUE} \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{proved}\:\mathrm{that}\:\frac{{d}}{{dx}}{f}\left({x}\right)\geqslant\mathrm{0}\:\Rightarrow\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{qx}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{q}}{{x}}}}\:\underset{\mathrm{0}<{x}\leqslant\mathrm{1}} {\leqslant}\:\:\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}+{q}}}. \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{A}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{B}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{C}}}\leqslant\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}+{q}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }}} \\ $$$$\mathrm{let}\:{g}\left({q}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}+{q}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{q}^{\mathrm{2}} }}} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\frac{{d}}{{dq}}{g}\left({q}\right)\underset{\mathrm{0}<{q}\leqslant\mathrm{1}} {\geqslant}\:\:\:\mathrm{0}. \\ $$$$\frac{{d}}{{dq}}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}+{q}}}+\frac{{q}}{\:\sqrt{\mathrm{1}+{q}^{\mathrm{2}} }}\right)= \\ $$$$\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}+{q}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} +\left(\mathrm{1}+{q}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} +{q}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}+{q}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \mathrm{2}{q}= \\ $$$$\left(\mathrm{1}+{q}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} −{q}^{\mathrm{2}} \left(\mathrm{1}+{q}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} −\left(\mathrm{1}+{q}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} = \\ $$$$\frac{\left(\mathrm{1}+{q}^{\mathrm{2}} \right)−{q}^{\mathrm{2}} }{\left(\mathrm{1}+{q}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+{q}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\frac{\mathrm{1}}{\left(\mathrm{1}+{q}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+{q}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\underset{\mathrm{0}<{q}\leqslant\mathrm{1}} {\geqslant}\:\:\:\mathrm{0}\:\Rightarrow \\ $$$$\underset{\mathrm{0}<{q}\leqslant\mathrm{1}} {\mathrm{max}}{g}\left({q}\right)={g}\left(\mathrm{1}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}+\mathrm{1}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{1}}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}. \\ $$$$ \\ $$

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