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Question Number 195124 by Shlock last updated on 25/Jul/23

Answered by witcher3 last updated on 25/Jul/23

(√x)+(√y)≤(√((x+1)(y+1))) ⇔x+y+2(√(xy))≤xy+x+y+1⇔xy+1≥2(√(xy)),AM−GM ⇒∀(x,y)∈R_+ (√x)+(√y)≤(√((x+1)(y+1))) ⇒∀(a,b)∈[1,∞[^2 (√(a−1))+(√(b−1))≤(√(ab)) (√(a−1))+(√(b−1))+(√(c−1))≤(√(ab))+(√(c−1))≤(√((ab+1)c))=(√(abc+c)) ⇔(√(a−1))+(√(b−1))+(√(c−1))≤(√(abc+c))

$$\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\leqslant\sqrt{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{y}+\mathrm{1}\right)} \\ $$$$\Leftrightarrow\mathrm{x}+\mathrm{y}+\mathrm{2}\sqrt{\mathrm{xy}}\leqslant\mathrm{xy}+\mathrm{x}+\mathrm{y}+\mathrm{1}\Leftrightarrow\mathrm{xy}+\mathrm{1}\geqslant\mathrm{2}\sqrt{\mathrm{xy}},\mathrm{AM}−\mathrm{GM} \\ $$$$\Rightarrow\forall\left(\mathrm{x},\mathrm{y}\right)\in\mathbb{R}_{+} \sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\leqslant\sqrt{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{y}+\mathrm{1}\right)} \\ $$$$\Rightarrow\forall\left(\mathrm{a},\mathrm{b}\right)\in\left[\mathrm{1},\infty\left[^{\mathrm{2}} \right.\right. \\ $$$$\sqrt{\mathrm{a}−\mathrm{1}}+\sqrt{\mathrm{b}−\mathrm{1}}\leqslant\sqrt{\mathrm{ab}} \\ $$$$\sqrt{\mathrm{a}−\mathrm{1}}+\sqrt{\mathrm{b}−\mathrm{1}}+\sqrt{\mathrm{c}−\mathrm{1}}\leqslant\sqrt{\mathrm{ab}}+\sqrt{\mathrm{c}−\mathrm{1}}\leqslant\sqrt{\left(\mathrm{ab}+\mathrm{1}\right)\mathrm{c}}=\sqrt{\mathrm{abc}+\mathrm{c}} \\ $$$$\Leftrightarrow\sqrt{\mathrm{a}−\mathrm{1}}+\sqrt{\mathrm{b}−\mathrm{1}}+\sqrt{\mathrm{c}−\mathrm{1}}\leqslant\sqrt{\mathrm{abc}+\mathrm{c}} \\ $$$$ \\ $$

Commented by Shlock last updated on 25/Jul/23

Perfect ��

Commented by witcher3 last updated on 26/Jul/23

thank You God bless You

$$\mathrm{thank}\:\mathrm{You}\:\mathrm{God}\:\mathrm{bless}\:\mathrm{You} \\ $$

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