Question

1.Cho tam giác ABC. Chứng minh:
\(\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\ge3\)
2. Cho x, y, z > 0 và xyz = 1. Tìm giá trị nhỏ nhất :
\(P=\frac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}+\frac{y^2\left(z+x\right)}{z\sqrt{z}+2x\sqrt{x}}+\frac{z^2\left(x+y\right)}{x\sqrt{x}+2y\sqrt{y}}\)

Answer 1

1) \(\left\{{}\begin{matrix}b+c-a=x\\c+a-b=y\\a+b-c=z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\frac{y+z}{2}\\b=\frac{z+x}{2}\\c=\frac{x+y}{2}\end{matrix}\right.\)

BĐT cần cm trở thành:

\(\frac{y+z}{2x}+\frac{z+x}{2y}+\frac{x+y}{2z}\ge3\)

Theo AM-GM, VT>=6/2=3

Dấu bằng xảy ra khi a=b=c

2)\(x^2\left(y+z\right)\ge2x^2\sqrt{yz}=2x^2\sqrt{\frac{1}{x}}=2x\sqrt{x}\)

=>\(P\ge\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}+\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)

Đặt \(\left\{{}\begin{matrix}x\sqrt{x}=a\\y\sqrt{y}=b\\z\sqrt{z}=c\end{matrix}\right.\Rightarrow abc=1\)

=>\(P\ge\frac{2a}{b+2c}+\frac{2b}{c+2a}+\frac{2c}{a+2b}\ge2.1=2\)

(Dùng Cauchy-Schwartz chứng minh được:

\(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\ge1\))

Dấu bằng xảy ra khi a=b=c=1 <=> x=y=z=1

Vậy minP=2<=>x=y=z=1