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f-x-1-1-2-2x-2x-x-4-prove-that-f-x-3-8-

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Question Number 79588 by loveineq. last updated on 26/Jan/20
f(x) = (1/(1+2(√(2x))+2x))+(x/4) prove that f(x) ≥ (3/8) .
$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}{x}}+\mathrm{2}{x}}+\frac{{x}}{\mathrm{4}} \\ $$$$\mathrm{prove}\:\mathrm{that}\:{f}\left({x}\right)\:\geqslant\:\frac{\mathrm{3}}{\mathrm{8}}\:. \\ $$
Commented by MJS last updated on 26/Jan/20
2(√(2x)) or 2(√2)x ?
$$\mathrm{2}\sqrt{\mathrm{2}{x}}\:\mathrm{or}\:\mathrm{2}\sqrt{\mathrm{2}}{x}\:? \\ $$
Commented by loveineq. last updated on 26/Jan/20
is 2(√(2x)) not 2(√2)x
$$\mathrm{is}\:\mathrm{2}\sqrt{\mathrm{2}{x}}\:\mathrm{not}\:\mathrm{2}\sqrt{\mathrm{2}}{x} \\ $$
Commented by john santu last updated on 26/Jan/20
let (√(2x)) = t ⇒2x = t^2 f(x) = (1/(1+2t+t^2 )) + (t^2 /8) f(x) = (1/((t+1)^2 )) +(t^2 /8) = (t+1)^(−2) +(t^2 /8) f ′(x) = {−2(t+1)^(−3) +(t/4)}×(1/t)=0 ((−2)/((t+1)^3 ))+(t/4)=0 ⇒t(t+1)^3 =8 t = 1 ⇒x = (1/2) ⇒f_(min) = (1/(1+2(√(2.(1/2)))+2.(1/2)))+(1/8) = (1/(1+2+1))+(1/8)=(3/8)
$$\mathrm{let}\:\sqrt{\mathrm{2x}}\:=\:\mathrm{t}\:\Rightarrow\mathrm{2x}\:=\:\mathrm{t}^{\mathrm{2}} \: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2t}+\mathrm{t}^{\mathrm{2}} }\:+\:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{8}}\:=\:\left(\mathrm{t}+\mathrm{1}\right)^{−\mathrm{2}} +\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)\:=\:\left\{−\mathrm{2}\left(\mathrm{t}+\mathrm{1}\right)^{−\mathrm{3}} +\frac{\mathrm{t}}{\mathrm{4}}\right\}×\frac{\mathrm{1}}{\mathrm{t}}=\mathrm{0} \\ $$$$\frac{−\mathrm{2}}{\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{t}}{\mathrm{4}}=\mathrm{0}\:\Rightarrow\mathrm{t}\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{8} \\ $$$$\mathrm{t}\:=\:\mathrm{1}\:\Rightarrow\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{f}_{\mathrm{min}} \:= \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}}+\mathrm{2}.\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$
Answered by MJS last updated on 26/Jan/20
(1/(2x+2(√2)x^(1/2) +1))+(x/4)<(3/8) ⇒ x^2 +(√2)x^(3/2) −x−((3(√2))/2)x^(1/2) +(5/4)<0 ((√x))^4 +(√2)((√x))^3 −((√x))^2 −((3(√2))/2)(√x)+(5/4)<0 ((√x)−((√2)/2))^2 (((√x))^2 +2(√2)(√x)+(5/2))<0 wrong ⇒ (1/(2x+2(√2)x^(1/2) +1))+(x/4)≥(3/8)
$$\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}\sqrt{\mathrm{2}}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{1}}+\frac{{x}}{\mathrm{4}}<\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} −{x}−\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} +\frac{\mathrm{5}}{\mathrm{4}}<\mathrm{0} \\ $$$$\left(\sqrt{{x}}\right)^{\mathrm{4}} +\sqrt{\mathrm{2}}\left(\sqrt{{x}}\right)^{\mathrm{3}} −\left(\sqrt{{x}}\right)^{\mathrm{2}} −\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{{x}}+\frac{\mathrm{5}}{\mathrm{4}}<\mathrm{0} \\ $$$$\left(\sqrt{{x}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} \left(\left(\sqrt{{x}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}\sqrt{{x}}+\frac{\mathrm{5}}{\mathrm{2}}\right)<\mathrm{0} \\ $$$$\mathrm{wrong}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{2}\sqrt{\mathrm{2}}{x}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{1}}+\frac{{x}}{\mathrm{4}}\geqslant\frac{\mathrm{3}}{\mathrm{8}} \\ $$

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