Cho x,y,z là các số thực thỏa mãn $xy+yz+zx=1$.Cmr:
$\frac{1}{x^2+2}$ + $\frac{1}{y^2+2}$+$\frac{1}{z^2+2}$ $\leq\frac{9}{7}$

Câu hỏi này được treo giải thưởng trị giá +5000 vỏ sò bởi laitridung2004@gmail.com, đã hết hạn vào lúc 10-03-19 03:10 PM

Nhân 2 vào 2 vế, ta có:
$\frac{2}{x^2+2}$+$\frac{2}{y^2+2}$+$\frac{2}{z^2+2}$ $\leq\frac{18}{7}$
Trừ mỗi vế đi 3, ta được 1 bất đẳng thức tương đương:
A=$\frac{x^2}{x^2+2}$+$\frac{y^2}{y^2+2}$+$\frac{z^2}{z^2+2}$ $\geq \frac{3}{7}$ (1)
Có:$3xy+yz+zx$ $\leq(x+y+z)^2$
Thật vậy, bất đẳng thức đã cho tương đương: $3xy+3yz+3zx \leq x^2+y^2+z^2+2xy+2yz+2zx$
$\Leftrightarrow$ $x^2+y^2+z^2-xy-yz-zx \geq 0$
$\Leftrightarrow$ $2x^2+2y^2+2z^2-2xy-2yz-2zx \geq 0$
$\Leftrightarrow$ $(x-y)^2+(y-z)^2+(z-x)^2 \geq 0$
Luôn đúng, nên $3(xy+yz+zx) \leq (x+y+z)^2 $
$\Leftrightarrow$ $3.4(xy+yz+zx)=3.4 \leq 4(x+y+z)^2 $
$\Leftrightarrow$ $4 \leq \frac{4(x+y+z)^2}{3} $
Quay trở lại bất đẳng thức (1): Áp dụng bất đẳng thức Bunhia-copxki ta có:
$A= \frac{x^2}{x^2+2} + \frac{y^2}{y^2+2} + \frac{z^2}{z^2+2}$
    $\geq \frac{(x+y+z)^2}{x^2+y^2+z^2+6}$
    $\geq \frac{(x+y+z)^2}{x^2+y^2+z^2+2xy+2yz+2zx+4}$
    $\geq \frac{(x+y+z)^2}{(x+y+z)^2+4}$
    $\geq \frac{(x+y+z)^2}{(x+y+z)^2 + \frac{4(x+y+z)^2}{3}}$
    $\geq \frac{3(x+y+z)^2}{7(x+y+z)^2}$
    $\geq \frac{3}{7}$ (Q.E.D)
Dấu bằng xảy ra $\Leftrightarrow x=y=z=\frac{1}{\sqrt3}$

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