Question Number 146817 by mathdanisur last updated on 15/Jul/21
Answered by mindispower last updated on 15/Jul/21
$$\Leftrightarrow\frac{{y}\left({x}+{y}\right)+{z}\left({x}+{z}\right)}{\:\sqrt{\left({x}+{y}\right)\left({x}+{z}\right)}}\geqslant{y}+{z}…. \\ $$$${y}\left({x}+{y}\right)+{z}\left({x}+{z}\right)={x}\left({y}+{z}\right)+{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \geqslant{x}\left({y}+{z}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left({y}+{z}\right)^{\mathrm{2}} \\ $$$$=\left({y}+{z}\right)\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\left({y}+{z}\right)\right)=\left({y}+{z}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\left({x}+{y}\right)+\left({x}+{z}\right)\right)\right) \\ $$$$\geqslant\left({y}+{z}\right).\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}\sqrt{\left({x}+{y}\right)\left({x}+{z}\right)}=\left({y}+{z}\right)\sqrt{\left({x}+{y}\right)\left({x}+{z}\right)} \\ $$$$\Rightarrow\frac{{y}\left({x}+{y}\right)+{z}\left({x}+{z}\right)}{\:\sqrt{\left({x}+{y}\right)\left({x}+{z}\right)}}\geqslant\frac{\left({y}+{z}\right)\sqrt{\left({x}+{y}\right)\left({x}+{z}\right)}}{\:\sqrt{\left({x}+{y}\right)\left({x}+{z}\right)}}={y}+{z} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mathdanisur last updated on 16/Jul/21
$${thank}\:{you}\:{Ser} \\ $$