Cho ba số thực dương thoả mãn $x^2+y^2+z^2+2xz=x+y+z$. Tìm Min $P=x+y+z+\dfrac{10}{\sqrt{x+y}}+\dfrac{10}{\sqrt{z+1}}$
Ta có : $\frac{1}{\sqrt{x+y}}+\frac{1}{\sqrt{z+1}}\geq \frac{2}{\sqrt[4]{(x+y)(z+1)}}\geq \frac{2\sqrt{2}}{\sqrt{x+y+z+1}}$
Do đó: $P\geq x+y+z+\frac{20\sqrt{2}}{\sqrt{x+y+z+1}}$
Theo giả thiết:
$ x^{2}+y^{2}+z^{2}+2xz=x+y+z$
$\Leftrightarrow (x+z)^{2}+y^{2}=x+y+z\\\Rightarrow x+y+z\geq \frac{(x+y+z)^{2}}{2}\\\Rightarrow 0< x+y+z\leq 2$
Xét $f(t)=t+\frac{20\sqrt{2}}{\sqrt{t+1}}$ $t\in \left (0;2 \right ] (t=x+y+z)$
$f'(t)< 0$ $\forall t\in \left (0;2 \right ]$
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