\(\ge\frac{3}{2}.\left(a+b+c\right)\) nhế mọi người, tui viết thiếu đề
Ta có bđt \(3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)(1)
Thật vậy\(\left(1\right)\Leftrightarrow3x^2+3y^2+3z^2\ge x^2+y^2+z^2+2xy+2yz+2zx\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)(Luôn đúng)
Áp dụng (1) và bđt Cô-si dạng engel\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)
\(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{9}{2a+2b+2c}\)
Nhân 2 vế bđt trên lại được
\(3\left(a^2+b^2+c^2\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\frac{9\left(a+b+c\right)^2}{2\left(a+b+c\right)}\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge\frac{3\left(a+b+c\right)}{2}\)
Dấu "=" <=> a=b=c
\(\left(a^2+b^2+c^2\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\frac{3}{2}\left(a+b+c\right)\left(1\right)\)
Ta có (1) <=> \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\). Dễ dàng suy ra từ BĐT Cosi
Áp dụng (1) ta được \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{9}{2\left(a+b+c\right)}\)
=> \(VT\ge\frac{9\left(a^2+b^2+c^2\right)}{2\left(a+b+c\right)}=\frac{3}{2}\cdot\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\ge\frac{3}{2}\left(a+b+c\right)\)
