cho $x,y,z>0$ thỏa mãn $(2x+y+z)(x+2y+z)(x+y+2z)=8$. Chứng minh rằng:
$\frac{x}{y(z+1)(y+z)^{2}}+\frac{y}{z(x+1)(z+x)^{2}}+\frac{z}{x(y+1(x+y)^{2}}\geq 2$
cho $x,y,z>0$ thỏa mãn $(2x+y+z)(x+2y+z)(x+y+2z)=8$. Chứng minh rằng:
$\frac{x}{y(z+1)(y+z)^{2}}+\frac{y}{z(x+1)(z+x)^{2}}+\frac{z}{x(y+1(x+y)^{2}}\geq 2$
Ta có bđt quen thuộc: $\prod (a+b)\geq \dfrac{8}{9}\sum a\sum ab$ $\forall a,b,c>0$
$$8=\prod (2x+y+z)=\prod \left[(x+y)+(z+x)\right]\geq \dfrac{8}{9}\left(2\sum x\right)\left[\sum (x+y)(y+z)\right]$$
$$=\dfrac{16}{9}\left(\sum x\right)\left(\sum x^{2}+3\sum xy\right)\geq \dfrac{16}{9}\sqrt{3\sum xy}\left(4\sum xy\right)$$
$$\Rightarrow \sqrt{\left(\sum xy\right)^{3}}\leq \dfrac{3\sqrt{3}}{8} \Rightarrow \sum xy\leq \dfrac{3}{4}$$
Mặt khác: $$3\sqrt[3]{\left(\prod x\right)^{2}}\leq \sum xy\leq \dfrac{3}{4} \Rightarrow \prod x\leq \dfrac{1}{8}$$
Trở lại bài toán: $$VT =\sum \dfrac{\left(\dfrac{x}{y+z}\right)^{2}}{xy(z+1)}\geq \dfrac{\left(\sum \dfrac{x}{y+z}\right)^{2}}{3\prod x+\sum xy}\geq \dfrac{\left(\dfrac{3}{2}\right)^{2}}{3.\dfrac{1}{8}+\dfrac{3}{4}}=2$$
Đẳng thức xảy ra $\Leftrightarrow x=y=z=\dfrac{1}{2}$
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