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 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 \("="\Leftrightarrow a=b=c\)

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 \(ab+bc+ca\le3\).

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

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

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

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 \(a=b=c\)

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

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

Dấu "=" xảy ra khi \(a=b=c\)

Với x dương, ta có đánh giá:

\(\dfrac{x}{1+x^2}\le\dfrac{36x+3}{50}\)

Thật vậy, BĐT tương đương:

\(\left(x^2+1\right)\left(36x+3\right)\ge50x\)

\(\Leftrightarrow36x^3+3x^2-14x+3\ge0\)

\(\Leftrightarrow\left(3x-1\right)^2\left(4x+3\right)\ge0\) (luôn đúng)

Áp dụng:

\(\dfrac{10a}{1+a^2}+\dfrac{10b}{1+b^2}+\dfrac{10c}{1+c^2}\le10.\dfrac{36\left(a+b+c\right)+9}{50}=9\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Sửa \(\le\) thành \(\ge\) nha bạn

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)

Ta có \(\dfrac{a^2}{a+bc}=\dfrac{a^3}{a^2+abc}=\dfrac{a^3}{a^2+ab+bc+ca}=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}\)

Tương tự: \(\left\{{}\begin{matrix}\dfrac{b^2}{b+ca}=\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}\\\dfrac{c^2}{c+ba}=\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\end{matrix}\right.\)

Áp dụng BĐT cosi:

\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3}{4}a\)

\(\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{b^3}{64}}=\dfrac{3}{4}b\)

\(\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{c^3}{64}}=\dfrac{3}{4}c\)

Cộng VTV:

\(\Leftrightarrow VT+\dfrac{a+b}{8}+\dfrac{a+c}{8}+\dfrac{b+c}{8}\ge\dfrac{3}{4}\left(a+b+c\right)\\ \Leftrightarrow VT\ge\dfrac{3\left(a+b+c\right)}{4}-\dfrac{2\left(a+b+c\right)}{8}\\ \Leftrightarrow VT\ge\dfrac{a+b+c}{4}\)

Dấu \("="\Leftrightarrow a=b=c=3\)

Lời giải:
$\text{VT}=\frac{a(a+b+c)+bc}{b+c}+\frac{b(a+b+c)+ac}{a+c}+\frac{c(a+b+c)+ab}{a+b}$
$=\frac{(a+b)(a+c)}{b+c}+\frac{(b+a)(b+c)}{a+c}+\frac{(c+a)(c+b)}{a+b}$

Áp dụng BĐT AM-GM:

$\frac{(a+b)(a+c)}{b+c}+\frac{(b+a)(b+c)}{a+c}\geq 2\sqrt{(a+b)^2}=2(a+b)$

$\frac{(b+c)(b+a)}{a+c}+\frac{(c+a)(c+b)}{a+b}\geq 2\sqrt{(b+c)^2}=2(b+c)$

$\frac{(a+b)(a+c)}{b+c}+\frac{(c+a)(c+b)}{a+b}\geq 2\sqrt{(c+a)^2}=2(a+c)$

Cộng các BĐT trên theo vế và thu gọn:

$\text{VT}\geq 2(a+b+c)=2$

Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

Đề sai. Bạn xem lại.