Question Number 151905 by mathdanisur last updated on 24/Aug/21
Answered by mindispower last updated on 24/Aug/21
![8x^x y^y z^z 2^(x+y+z) =2(2x)^x .2(2y)^y .(2z)^z_ 2(2t)^t ≥(t+1)^(t+1) ,t>0 ⇔2^(t+1) ≥t(1+(1/t))^(t+1) ⇔(t+1)ln(2)≥ln(t)+(t+1)ln(1+(1/t)) f(t)=(t+1)ln(2)−ln(t)−(t+1)ln(1+(1/t)) f′(t)=ln(2)−(1/t)−ln(1+(1/t))+(1/t)=ln(((2t)/(t+1))) f′(t)≥0,t∈[1,+∞[^ ,f′(t)<0,t∈]0,1[ ⇒f(t)≥f(1)=0 ⇒∀t>0,2.(2t)^(t+1) ≥(t+1)^(t+1) ⇒2.(2x)^(x+1) .2(2y)^(y+1) .2(2z)^(z+1) ≥(x+1)^(x+1) (y+1)^(y+1) (z+1)^(z+1) ⇔8x^x .y^y .z^z .2^(x+y+z) ≥(x+1)^(x+1) (y+1)^(y+1) (z+1)^(z+1)](https://www.tinkutara.com/question/Q151939.png)
$$\mathrm{8}{x}^{{x}} {y}^{{y}} {z}^{{z}} \mathrm{2}^{{x}+{y}+{z}} =\mathrm{2}\left(\mathrm{2}{x}\right)^{{x}} .\mathrm{2}\left(\mathrm{2}{y}\right)^{{y}} .\left(\mathrm{2}{z}\right)^{{z}_{} } \\ $$$$\mathrm{2}\left(\mathrm{2}{t}\right)^{{t}} \geqslant\left({t}+\mathrm{1}\right)^{{t}+\mathrm{1}} ,{t}>\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{2}^{{t}+\mathrm{1}} \geqslant{t}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)^{{t}+\mathrm{1}} \\ $$$$\Leftrightarrow\left({t}+\mathrm{1}\right){ln}\left(\mathrm{2}\right)\geqslant{ln}\left({t}\right)+\left({t}+\mathrm{1}\right){ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right) \\ $$$${f}\left({t}\right)=\left({t}+\mathrm{1}\right){ln}\left(\mathrm{2}\right)−{ln}\left({t}\right)−\left({t}+\mathrm{1}\right){ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right) \\ $$$${f}'\left({t}\right)={ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{{t}}−{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)+\frac{\mathrm{1}}{{t}}={ln}\left(\frac{\mathrm{2}{t}}{{t}+\mathrm{1}}\right) \\ $$$${f}'\left({t}\right)\geqslant\mathrm{0},{t}\in\left[\mathrm{1},+\infty\hat {\left[},{f}'\left({t}\right)<\mathrm{0},{t}\in\right]\mathrm{0},\mathrm{1}\left[\right.\right. \\ $$$$\Rightarrow{f}\left({t}\right)\geqslant{f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\forall{t}>\mathrm{0},\mathrm{2}.\left(\mathrm{2}{t}\right)^{{t}+\mathrm{1}} \geqslant\left({t}+\mathrm{1}\right)^{{t}+\mathrm{1}} \\ $$$$\:\Rightarrow\mathrm{2}.\left(\mathrm{2}{x}\right)^{{x}+\mathrm{1}} .\mathrm{2}\left(\mathrm{2}{y}\right)^{{y}+\mathrm{1}} .\mathrm{2}\left(\mathrm{2}{z}\right)^{{z}+\mathrm{1}} \geqslant\left({x}+\mathrm{1}\right)^{{x}+\mathrm{1}} \left({y}+\mathrm{1}\right)^{{y}+\mathrm{1}} \left({z}+\mathrm{1}\right)^{{z}+\mathrm{1}} \\ $$$$\Leftrightarrow\mathrm{8}{x}^{{x}} .{y}^{{y}} .{z}^{{z}} .\mathrm{2}^{{x}+{y}+{z}} \geqslant\left({x}+\mathrm{1}\right)^{{x}+\mathrm{1}} \left({y}+\mathrm{1}\right)^{{y}+\mathrm{1}} \left({z}+\mathrm{1}\right)^{{z}+\mathrm{1}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mathdanisur last updated on 24/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$
Commented by mindispower last updated on 28/Aug/21
$${plrasur} \\ $$