Ta có : \(\left(a^2+1\right)\left(b^2+1\right)=\left(a^2+ab+a+b\right)\left(b^2+ab+a+b\right)\)

\(=\left(a+1\right)\left(a+b\right)\left(b+1\right)\left(a+b\right)=\left(ab+a+b+1\right)\left(a+b\right)^2\)

\(=\left(1+1\right)\left(a+b\right)^2=2\left(a+b\right)^2\)(đpcm)

chịu ai bt đc 90% là 2k10 mà

Ta có:\(\left(a^2+1\right)\left(b^2+1\right)=2\left(a+b\right)^2\)

Xét \(VT=\left(a^2+1\right)\left(b^2+1\right)\)

\(=\left(a^2+ab+a+b\right)\left(b^2+ab+a+b\right)\)Do \(ab+a+b=1\)

\(=\left[a\left(a+b\right)+\left(a+b\right)\right]\left[b\left(a+b\right)+\left(a+b\right)\right]\)

\(=\left(a+b\right)\left(a+1\right)\left(b+1\right)\left(a+b\right)\)

\(=\left(a+b\right)^2\left(ab+a+b+1\right)\)Do \(ab+a+b=1\)

\(=\left(a+b\right)^2\left(1+1\right)\)

\(=2\left(a+b\right)^2=VP\)

1)Cho a,b,c >0

Chứng minh  bc/a^2(b+c) + ca/b^2(c+a) +ab/c^2(a+b) > hoặc = 1/2(1/a+1/b+1/c)

2) Cho a,b,c>0 1/a + 1/b + 1/c =1

Chứng minh (b+c)/a^2 + (c+a)/b^2 + (a+b)/c^2 > hoặc = 2

Đọc tiếp...

Lời giải:
BĐT \(\Leftrightarrow \frac{a^2+b^2+2}{(a^2+1)(b^2+1)}\geq \frac{2}{ab+1}\)

$\Leftrightarrow (a^2+b^2+2)(ab+1)\geq 2(a^2b^2+a^2+b^2+1)$

$\Leftrightarrow a^3b+a^2+ab^3+b^2+2ab+2\geq 2a^2b^2+2a^2+2b^2+2$

$\Leftrightarrow a^3b+ab^3+2ab\geq 2a^2b^2+a^2+b^2$

$\Leftrightarrow ab(a^2+b^2-2ab)-(a^2+b^2-2ab)\geq 0$

$\Leftrightarrow ab(a-b)^2-(a-b)^2\geq 0$

$\Leftrightarrow (a-b)^2(ab-1)\geq 0$

Điều này luôn đúng với mọi $ab\geq 1$ 

Do đó ta có đpcm 

Dấu "=" xảy ra khi $a=b$ hoặc $ab=1$