\(\Leftrightarrow\left(2+a^2+b^2\right)\left(1+ab\right)\ge2\left(1+a^2\right)\left(1+b^2\right)\)

\(\Leftrightarrow2+2ab+a^2+b^2+ab\left(a^2+b^2\right)\ge2+2a^2+2b^2+2a^2b^2\)

\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\)

\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\) (luôn đúng với mọi \(a\ge1;b\ge1\))

Cách khác:

\(\Leftrightarrow\left(\frac{1}{1+a^2}-\frac{1}{1+ab}\right)+\left(\frac{1}{1+b^2}-\frac{1}{1+ab}\right)\ge0\)

\(\Leftrightarrow\frac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}+\frac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)\left[b\left(1+a^2\right)-a\left(1+b^2\right)\right]}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2\left(ab-1\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\) (luôn đúng).

\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\Leftrightarrow\frac{2+a^2+b^2}{\left(1+a^2+b^2+a^2b^2\right)}\ge\frac{2}{1+ab}\)

\(\Leftrightarrow\left(1+ab\right)\left(2+a^2+b^2\right)\ge2a^2b^2+2a^2+2b^2+2\)

\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\)

\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)

b/ \(\frac{1}{1+a^4}+\frac{1}{1+b^4}+\frac{2}{1+b^4}\ge\frac{2}{1+a^2b^2}+\frac{2}{1+b^4}\ge\frac{4}{1+ab^3}\)

\(\Rightarrow\frac{1}{1+a^4}+\frac{3}{1+b^4}\ge\frac{4}{1+ab^3}\)

Hoàn toàn tương tự: \(\frac{1}{1+b^4}+\frac{3}{1+c^4}\ge\frac{4}{1+bc^3}\); \(\frac{1}{1+c^4}+\frac{3}{1+a^4}\ge\frac{4}{1+a^3c}\)

Cộng vế với vế ta có đpcm

khó quá bạn ơi mình cần thêm thời gian để làm 

nhanh lên nhé

Chứng minh bằng biến đổi tương đương : 

\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)

\(\Leftrightarrow\left(\frac{1}{1+a^2}-\frac{1}{1+ab}\right)+\left(\frac{1}{1+b^2}-\frac{1}{1+ab}\right)\ge0\)

\(\Leftrightarrow\frac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}+\frac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Leftrightarrow\left(\frac{a-b}{1+ab}\right)\left(\frac{b}{1+b^2}-\frac{a}{1+a^2}\right)\ge0\)

\(\Leftrightarrow\frac{a-b}{1+ab}.\frac{\left(a-b\right)\left(ab-1\right)}{\left(1+a^2\right)\left(1+b^2\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2\left(ab-1\right)}{\left(ab+1\right)\left(a^2+1\right)\left(b^2+1\right)}\ge0\)

Vì \(a\ge1,b\ge1\) nên \(ab-1\ge0\) . Mặt khác vì \(\left(a-b\right)^2\ge0\) nên ta có điều phải chứng minh.

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Ta có: \(a\ge b\Rightarrow1+b^2\le1+a^2\)

\(\Rightarrow\frac{1}{1+b^2}\ge\frac{1}{1+a^2}\Rightarrow\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{1}{1+a^2}+\frac{1}{1+a^2}\)

\(\Leftrightarrow\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+a^2}\)

Lời giải:

Biến đổi tương đương:
\(\frac{1}{x^2+1}+\frac{1}{y^2+1}\geq \frac{2}{1+xy}\)

\(\Leftrightarrow \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)}\geq \frac{2}{xy+1}\)

\(\Leftrightarrow (xy+1)(x^2+y^2+2)\geq 2(x^2+1)(y^2+1)\)

\(\Leftrightarrow xy(x^2+y^2)+2xy+x^2+y^2+2\geq 2x^2y^2+2x^2+2y^2+2\)

\(\Leftrightarrow xy(x^2+y^2)+2xy-2x^2y^2-x^2-y^2\geq 0\)

\(\Leftrightarrow xy(x^2+y^2-2xy)-(x^2-2xy+y^2)\geq 0\)

\(\Leftrightarrow xy(x-y)^2-(x-y)^2\geq 0\leftrightarrow (xy-1)(x-y)^2\geq 0\)

BĐT trên luôn đúng với mọi $x\geq 1, y\geq 1$. Do đó ta có đpcm.

Dấu "=" xảy ra khi $xy=1$ hoặc $x=y\geq 1$

\(1.\sqrt{a^2+ab+b^2}\le\frac{1+a^2+ab+b^2}{2}\)

\(\Rightarrow VT\ge\frac{1}{\frac{1+a^2+ab+b^2}{2}}+\)\(\frac{1}{\frac{1+b^2+cb+c^2}{2}}+\)\(\frac{1}{\frac{1+c^2+ac+a^2}{2}}\)\(\ge\frac{\left(1+1+1\right)^2}{\frac{1+a^2+ab+b^2}{2}+\frac{1+b^2+bc+c^2}{2}+\frac{1+c^2+ca+a^2}{2}}=\frac{9}{a^2+b^2+c^2+\frac{\left(ab+bc+ca\right)+3}{2}}\ge\frac{9}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}=VP\)

vì   3 </ 3 ( ab+bc+ca)