This pro will be solved by SOS method

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

\(\Leftrightarrow\dfrac{a^2b+a^2c-ab^2-ac^2}{\left(b^2+c^2\right)\left(b+c\right)}+\dfrac{ab^2+b^2c-a^2b-bc^2}{\left(a^2+c^2\right)\left(a+c\right)}+\dfrac{bc^2+ac^2-a^2c-b^2c}{\left(a^2+b^2\right)\left(a+b\right)}\ge0\)

\(\Leftrightarrow\dfrac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b^2+c^2\right)\left(b+c\right)}+\dfrac{ab\left(b-a\right)+bc\left(c-b\right)}{\left(a^2+c^2\right)\left(a+c\right)}+\dfrac{bc\left(c-b\right)+ac\left(c-a\right)}{\left(a^2+b^2\right)\left(a+b\right)}\ge0\)

\(\LeftrightarrowΣ\left(\dfrac{ab\left(a-b\right)}{\left(b^2+c^2\right)\left(b+c\right)}+\dfrac{ab\left(b-a\right)}{\left(a^2+c^2\right)\left(a+c\right)}\right)\ge0\)

\(\LeftrightarrowΣ\left(ab\left(a-b\right)\left(\dfrac{\left(a^2+c^2\right)\left(a+c\right)-\left(b^2+c^2\right)\left(b+c\right)}{\left(b^2+c^2\right)\left(a^2+c^2\right)\left(b+c\right)\left(a+c\right)}\right)\right)\ge0\)

\(\LeftrightarrowΣ\left(ab\left(a-b\right)\cdot\dfrac{\left(a-b\right)\left(a^2+b^2+c^2+ab+bc+ca\right)}{\left(b^2+c^2\right)\left(a^2+c^2\right)\left(b+c\right)\left(a+c\right)}\right)\ge0\)

\(\LeftrightarrowΣ\left(ab\left(a-b\right)^2\dfrac{a^2+b^2+c^2+ab+bc+ca}{\left(b^2+c^2\right)\left(a^2+c^2\right)\left(b+c\right)\left(a+c\right)}\right)\ge0\) *RIGHT!*

a 2 b 2 + c 2+ b 2 c 2 + a 2+ c 2 a 2 + b 2≥ a b + c+ b c + a+ c a + ba2b2+c2+b2c2+a2+c2a2+b2≥ab+c+bc+a+ca+b

⇔ a 2 b + a 2 c - a b 2 - a c 2 ( b 2 + c 2 ) ( b + c )+ Một b 2 + b 2 c - một 2 b - b c 2 ( một 2 + c 2 ) ( một + c )+ b c 2 + a c 2 - a 2 c - b 2 c ( a 2 + b 2 ) ( a + b )≥ 0⇔a2b+a2c−ab2−ac2(b2+c2)(b+c)+ab2+b2c−a2b−bc2(a2+c2)(a+c)+bc2+ac2−a2c−b2c(a2+b2)(a+b)≥0

⇔ a b ( a - b ) + a c ( a - c ) ( b 2 + c 2 ) ( b + c )+ Một b ( b - một ) + b c ( c - b ) ( một 2 + c 2 ) ( một + c )+ b c ( c - b ) + a c ( c - a ) ( a 2 + b 2 ) ( a + b )≥ 0⇔ab(a−b)+ac(a−c)(b2+c2)(b+c)+ab(b−a)+bc(c−b)(a2+c2)(a+c)+bc(c−b)+ac(c−a)(a2+b2)(a+b)≥0

⇔ Σ ( a b ( a - b ) ( b 2 + c 2 ) ( b + c )+ Một b ( b - một ) ( một 2 + c 2 ) ( một + c )) ≥ 0⇔Σ(ab(a−b)(b2+c2)(b+c)+ab(b−a)(a2+c2)(a+c))≥0

⇔ Σ ( một b ( một - b ) ( ( một 2 + c 2 ) ( một + c ) - ( b 2 + c 2 ) ( b + c ) ( b 2 + c 2 ) ( một 2 + c 2 ) ( b + c ) ( a + c )) )≥0⇔Σ(ab(a−b)((a2+c2)(a+c)−(b2+c2)(b+c)(b2+c2)(a2+c2)(b+c)(a+c)))≥0

⇔ Σ ( một b ( một - b ) ⋅ ( một - b ) ( một 2 + b 2 + c 2 + một b + b c + c một ) ( b 2 + c 2 ) ( một 2 + c 2 ) ( b + c ) ( a + c )) ≥ 0⇔Σ(ab(a−b)⋅(a−b)(a2+b2+c2+ab+bc+ca)(b2+c2)(a2+c2)(b+c)(a+c))≥0

⇔ Σ ( một b ( một - b ) 2 một 2 + b 2 + c 2 + một b + b c + c một ( b 2 + c 2 ) ( một 2 + c 2 ) ( b + c ) ( một + c )

This pro will be solved by SOS method

a2b2+c2+b2c2+a2+c2a2+b2≥ab+c+bc+a+ca+ba2b2+c2+b2c2+a2+c2a2+b2≥ab+c+bc+a+ca+b

⇔a2b+a2c−ab2−ac2(b2+c2)(b+c)+ab2+b2c−a2b−bc2(a2+c2)(a+c)+bc2+ac2−a2c−b2c(a2+b2)(a+b)≥0⇔a2b+a2c−ab2−ac2(b2+c2)(b+c)+ab2+b2c−a2b−bc2(a2+c2)(a+c)+bc2+ac2−a2c−b2c(a2+b2)(a+b)≥0

⇔ab(a−b)+ac(a−c)(b2+c2)(b+c)+ab(b−a)+bc(c−b)(a2+c2)(a+c)+bc(c−b)+ac(c−a)(a2+b2)(a+b)≥0⇔ab(a−b)+ac(a−c)(b2+c2)(b+c)+ab(b−a)+bc(c−b)(a2+c2)(a+c)+bc(c−b)+ac(c−a)(a2+b2)(a+b)≥0

⇔Σ(ab(a−b)(b2+c2)(b+c)+ab(b−a)(a2+c2)(a+c))≥0⇔Σ(ab(a−b)(b2+c2)(b+c)+ab(b−a)(a2+c2)(a+c))≥0

⇔Σ(ab(a−b)((a2+c2)(a+c)−(b2+c2)(b+c)(b2+c2)(a2+c2)(b+c)(a+c)))≥0⇔Σ(ab(a−b)((a2+c2)(a+c)−(b2+c2)(b+c)(b2+c2)(a2+c2)(b+c)(a+c)))≥0

⇔Σ(ab(a−b)⋅(a−b)(a2+b2+c2+ab+bc+ca)(b2+c2)(a2+c2)(b+c)(a+c))≥0⇔Σ(ab(a−b)⋅(a−b)(a2+b2+c2+ab+bc+ca)(b2+c2)(a2+c2)(b+c)(a+c))≥0

⇔Σ(ab(a−b)2a2+b2+c2+ab+bc+ca(b2+c2)(a2+c2)(b+c)(a+c))≥0⇔Σ(ab(a−b)2a2+b2+c2+ab+bc+ca(b2+c2)(a2+c2)(b+c)(a+c))≥0 *RIGHT!*

Alchemy I can't answer my question and I'm lazy

ok you can post your solution which you think smaller than I =))

Alchemy I have a answer quicker than your answer but I want to a answer quickest.

A compact answer ? You need to know that this SOS method is one of method COMPACT because it doesn't loss of variables. 

Alchemy Your answer is right but I need a compact answer