Sử dụng AM-GM trong chứng minh BĐT.

Question

1. Cho $x,\,y,\,z>0$ và $xyz=1.$ Chứng minh rằng: $\dfrac{x^2}{1+y}+\dfrac{y^2}{1+z}+\dfrac{z^2}{1+x}\geq\dfrac{3}{2}$

2. Cho $x,\,y,\,z>0$ thỏa mãn $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=4.$ Chứng minh rằng: $\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\leq1$

score 6 · Answer 1

2.
Áp dụng BĐT Cauchy ta có:
$(2x+y+z)(\frac{2}{x}+\frac{1}{y}+\frac{1}{z})=(x+x+y+z)(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$
$\ge4\sqrt[4]{x^2yz}.4\sqrt[4]{\frac{1}{x^2yz}}=16$
Suy ra: $\frac{1}{2x+y+z}\le\frac{1}{16}(\frac{2}{x}+\frac{1}{y}+\frac{1}{z})$
Tương tự: $\frac{1}{x+2y+z}\le\frac{1}{16}(\frac{1}{x}+\frac{2}{y}+\frac{1}{z})$
$\frac{1}{x+y+2z}\le\frac{1}{16}(\frac{1}{x}+\frac{1}{y}+\frac{2}{z})$
Suy ra: $\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{16}(\frac{4}{x}+\frac{4}{y}+\frac{4}{z})=1$
Dấu bằng xảy ra khi: $x=y=z=\frac{3}{4}$

score 6 · Answer 2

Bình chọn tăng 6 Bình chọn giảm

1.
Áp dụng BĐT Cauchy ta có:
$\frac{x^2}{1+y}+\frac{1+y}{4}\ge2\sqrt{\frac{x^2}{1+y}.\frac{1+y}{4}}=x$
$\frac{y^2}{1+z}+\frac{1+z}{4}\ge2\sqrt{\frac{y^2}{1+z}.\frac{1+z}{4}}=y$
$\frac{z^2}{1+x}+\frac{1+x}{4}\ge2\sqrt{\frac{z^2}{1+x}.\frac{1+x}{4}}=z$
Cộng các BĐT trên ta có:
$\frac{x^2}{1+y}+\frac{y^2}{1+z}+\frac{z^2}{1+x}\ge\frac{3}{4}(x+y+z)-\frac{3}{4}$
$\ge\frac{9}{4}\sqrt[3]{xyz}-\frac{3}{4}=\frac{3}{2}$
Dấu bằng xảy ra khi: $x=y=z=1$