Ta có: \(\dfrac{a^2}{b^2}+1\ge2\sqrt{\dfrac{a^2}{b^2}}=\dfrac{2a}{b}\)

Tương tự: \(\dfrac{b^2}{c^2}+1\ge\dfrac{2b}{c}\) ; \(\dfrac{c^2}{a^2}+1\ge\dfrac{2c}{a}\)

\(\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+3\ge\dfrac{2a}{b}+\dfrac{2b}{c}+\dfrac{2c}{a}\) (1)

Mà \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge3\sqrt[3]{\dfrac{abc}{abc}}=3\)

\(\Rightarrow\dfrac{2a}{b}+\dfrac{2b}{c}+\dfrac{2c}{a}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+3\) (2)

(1);(2) \(\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+3\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+3\)

\(\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)

Dấu "=" xảy ra khi \(a=b=c\)

\(VT=\left(\dfrac{b}{a}+\dfrac{b}{c}\right)+\left(\dfrac{c}{a}+\dfrac{c}{b}\right)+\left(\dfrac{a}{b}+\dfrac{a}{c}\right)\)

Ta có \(\left(\dfrac{b}{c}+\dfrac{b}{a}\right)\left(a+c\right)\ge\left(\sqrt{b}+\sqrt{b}\right)^2=4b\Leftrightarrow\dfrac{b}{c}+\dfrac{b}{a}\ge\dfrac{4b}{a+c}\)

CMTT \(\Leftrightarrow\left(\dfrac{c}{a}+\dfrac{c}{b}\right)\ge\dfrac{4c}{a+b};\dfrac{a}{b}+\dfrac{a}{c}\ge\dfrac{4a}{b+c}\)

Cộng VTV ta đc đpcm

Dấu \("="\Leftrightarrow a=b=c\)

Lời giải:
$\text{VT}=\sum \frac{a^2}{a+b^2}=\sum (a-\frac{ab^2}{a+b^2})$

$=\sum a-\sum \frac{ab^2}{a+b^2}$

$\geq \sum a-\sum \frac{ab^2}{2b\sqrt{a}}$ (theo BĐT AM-GM)

$=\sum a-\frac{1}{2}\sum \sqrt{ab^2}$

$\geq \sum a-\frac{1}{2}\sum \frac{ab+b}{2}$ (AM-GM)

$=\frac{3}{4}\sum a-\frac{1}{4}\sum ab$

Giờ ta chỉ cần cm $\sum a\geq \sum ab$ là bài toán được giải quyết

Thật vậy:
Đặt $\sum ab=t$ thì hiển nhiên $0< t\leq 3$ theo BĐT AM-GM

$(\sum a)^2-(\sum ab)^2=3+2t-t^2=(3-t)(t+1)\geq 0$ với mọi $0< t\leq 3$

$\Rightarrow \sum a\geq \sum ab$

Vậy ta có đcpcm.

Dấu "=" xảy ra khi $a=b=c$

Áp dụng bất đẳng thức Chevbyshev cho hai bộ đơn điệu cùng chiều \(\left(\dfrac{2}{a+b},\dfrac{2}{b+c},\dfrac{2}{c+a}\right)\) và \(\left(c\left(a+b\right),a\left(b+c\right),b\left(c+a\right)\right)\) ta có \(2c+2a+2b=\dfrac{2}{a+b}.c\left(a+b\right)+\dfrac{2}{b+c}.a\left(b+c\right)+\dfrac{2}{c+a}.b\left(c+a\right)\ge\dfrac{1}{3}\left(\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\right)\left(a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)\right)=\dfrac{2}{3}\left(ab+bc+ca\right)\left(\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\right)\).

Mà \(\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}=a+b+c\) nên \(ab+bc+ca\le3\).

Bài này có bạn giải rồi:

Cho các số thực dương a,b,c.Chứng minh rằng :\(\dfrac{b\left(2a-b\right)}{a\left(b+c\right)}+\dfrac{c\left(2b-c\right)}{... - Hoc24

\(VT=\dfrac{\left(a+c\right)^2}{\left(a+c\right)\left(a+b\right)}+\dfrac{\left(b+d\right)^2}{\left(b+c\right)\left(b+d\right)}+\dfrac{\left(c+a\right)^2}{\left(c+a\right)\left(c+d\right)}+\dfrac{\left(d+b\right)^2}{\left(d+a\right)\left(d+b\right)}\)

\(VT\ge\dfrac{\left(2a+2b+2c+2d\right)^2}{\left(a+b\right)\left(a+c\right)+\left(b+c\right)\left(b+d\right)+\left(a+c\right)\left(c+d\right)+\left(a+d\right)\left(b+d\right)}=\dfrac{4\left(a+b+c+d\right)^2}{\left(a+b+c+d\right)^2}=4\)

Dấu "=" xảy ra khi \(a=b=c=d\)

\(\left(a+b+c\right)\left(\dfrac{a}{\left(b+c\right)^2}+\dfrac{b}{\left(c+a\right)^2}+\dfrac{c}{\left(a+b\right)^2}\right)\ge\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\ge\dfrac{9}{4}\)

\(\Rightarrow\dfrac{a}{\left(b+c\right)^2}+\dfrac{b}{\left(c+a\right)^2}+\dfrac{c}{\left(a+b\right)^2}\ge\dfrac{9}{4\left(a+b+c\right)}\)

Dấu "=" xảy ra khi \(a=b=c\)

Áp dụng BĐT Cosi:

\(\dfrac{a}{\sqrt{b^2+ab}}=\dfrac{a\sqrt{2}}{\sqrt{2\left(b^2+ab\right)}}=\dfrac{a\sqrt{2}}{\sqrt{2b\left(a+b\right)}}\ge\dfrac{a\sqrt{2}}{\dfrac{2b+a+b}{2}}=\dfrac{2\sqrt{2}a}{a+3b}\)

Cmtt: \(\dfrac{b}{\sqrt{c^2+bc}}\ge\dfrac{2\sqrt{2}b}{b+3c};\dfrac{c}{\sqrt{a^2+ca}}\ge\dfrac{2\sqrt{2}c}{c+3a}\)

\(\Leftrightarrow P\ge2\sqrt{2}\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)\\ \Leftrightarrow\dfrac{P}{\sqrt{2}}\ge2\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)\\ \Leftrightarrow\dfrac{P}{\sqrt{2}}\ge\dfrac{2\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\dfrac{1}{3}\left(a+b+c\right)^2}\\ \Leftrightarrow\dfrac{P}{\sqrt{2}}\ge\dfrac{2}{\dfrac{4}{3}}=\dfrac{3}{2}\\ \Leftrightarrow P\ge\dfrac{3\sqrt{2}}{2}\)

Dấu \("="\Leftrightarrow a=b=c\)

\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge3+\dfrac{2a^2+2b^2+2c^2-2\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\)

\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}\ge5-\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\)

\(\Leftrightarrow\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\ge5\)

Do \(\dfrac{2a}{b+c}+\dfrac{2b}{c+a}+\dfrac{2c}{a+b}=\dfrac{2a^2}{ab+ac}+\dfrac{2b^2}{bc+ab}+\dfrac{2c^2}{ac+bc}\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}\)

Nên ta chỉ cần chứng minh:

\(\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\ge5\)

Điều này hiển nhiên đúng do:

\(VT=\dfrac{2}{3}.\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{6\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}+\dfrac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\)

\(VT\ge2\sqrt{\dfrac{12\left(a+b+c\right)^2\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)\left(a+b+c\right)^2}}+\dfrac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=5\)

Dấu "=" xảy ra khi \(a=b=c\)

\(\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}\ge2\)

\(\Leftrightarrow\dfrac{2}{a+2}-1+\dfrac{2}{b+2}-1+\dfrac{2}{c+2}-1\ge2-3\)

\(\Rightarrow1\ge\dfrac{a}{a+2}+\dfrac{b}{b+2}+\dfrac{c}{c+2}=\dfrac{a^2}{a^2+2a}+\dfrac{b^2}{b^2+2b}+\dfrac{c^2}{c^2+2c}\)

\(\Rightarrow1\ge\dfrac{\left(a+b+c\right)^2}{a^2+2a+b^2+2b+c^2+2c}\)

\(\Rightarrow a^2+b^2+c^2+2\left(a+b+c\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Rightarrow\) đpcm

Phía trên thoả mãn \(\ge1\) chứ không phải 3/2 đâu ạ