Question

Cho x, y, z > 0 thỏa mãn : xy + yz + xz = 3.

CMR : \(\frac{1}{x^2+y^2+2}+\frac{1}{y^2+z^2+2}+\frac{1}{z^2+x^2+2}\le\frac{3}{4}\)

Answer 1

Lời giải:

Ta có:

\(\text{VT}=\frac{1}{x^2+y^2+2}+\frac{1}{y^2+z^2+2}+\frac{1}{z^2+x^2+2}\)

\(\Rightarrow 2\text{VT}=\frac{2}{x^2+y^2+2}+\frac{2}{y^2+z^2+2}+\frac{2}{z^2+x^2+2}\)

\(2\text{VT}=1-\frac{x^2+y^2}{x^2+y^2+2}+1-\frac{y^2+z^2}{y^2+z^2+2}+1-\frac{z^2+x^2}{z^2+x^2+2}\)

\(2\text{VT}=3-\left(\frac{x^2+y^2}{x^2+y^2+2}+\frac{y^2+z^2}{y^2+z^2+2}+\frac{z^2+x^2}{z^2+x^2+2}\right)=3-A\)

Áp dụng BĐT Cauchy-Schwarz:

\(A\geq \frac{(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2}{2(x^2+y^2+z^2)+6}=\frac{(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2}{2(x^2+y^2+z^2+xy+yz+xz)}(*)\)

Xét tử số:

\((\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2\)

\(=2(x^2+y^2+z^2)+2(\sqrt{(x^2+y^2)(x^2+z^2)}+\sqrt{(x^2+y^2)(y^2+z^2)}+\sqrt{(y^2+z^2)(z^2+x^2)})\)

Áp dụng BĐT Bunhiacopxky:

\(\sqrt{(x^2+y^2)(x^2+z^2)}\geq \sqrt{(x^2+yz)^2}=x^2+yz\)

\(\sqrt{(x^2+y^2)(y^2+z^2)}\geq \sqrt{(xz+y^2)^2}=xz+y^2\)

\(\sqrt{(y^2+z^2)(z^2+x^2)}\geq \sqrt{(z^2+xy)^2}=z^2+xy\)

\(\Rightarrow \sum \sqrt{(x^2+y^2)(x^2+z^2)}\geq x^2+y^2+z^2+xy+yz+xz\)

\(\Rightarrow (\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2\geq 4(x^2+y^2+z^2)+2(xy+yz+xz)\)

\(\geq 3(x^2+y^2+z^2)+3(xy+yz+xz)=3(x^2+y^2+z^2+xy+yz+xz)\)

(theo BĐT AM-GM)

Do đó: Từ \((*)\Rightarrow A\geq \frac{3(x^2+y^2+z^2+xy+yz+xz)}{2(x^2+y^2+z^2+xy+yz+xz)}=\frac{3}{2}\)

\(\Rightarrow 2\text{VT}\leq 3-\frac{3}{2}=\frac{3}{2}\)

\(\Rightarrow \text{VT}\leq \frac{3}{4}\) (đpcm)

Dấu bằng xảy ra khi \(x=y=z=1\)

Answer 2

We have: \(\dfrac{1}{x^2+y^2+2}=\dfrac{1}{x^2+y^2+z^2+2-z^2}\le\dfrac{1}{5-z^2}\)

Similarly and by adding them:

\(\dfrac{1}{5-x^2}+\dfrac{1}{5-y^2}+\dfrac{1}{5-z^2}\le\dfrac{3}{4}\left(\circledast\right)\)

We know that \(\dfrac{1}{5-x^2}\le\dfrac{3\left(x^2+x\right)}{8\left(x^2+x+1\right)}\)

\(\Leftrightarrow-\dfrac{\left(x-1\right)^2\left(3x^2+9x+8\right)}{8\left(x^2-5\right)\left(x^2+x+1\right)}\le0\) It's obviously

\(\Rightarrow L.H.S_{\left(\circledast\right)}\le\dfrac{3}{8}\left(\dfrac{x^2+x}{x^2+x+1}+\dfrac{y^2+y}{y^2+y+1}+\dfrac{z^2+z}{z^2+z+1}\right)\le\dfrac{3}{4}\)

The equality occur when \(x=y=z=1\)

Done!