Áp dụng AM-GM:
\(a^5+1+1+1+1\ge5a\Rightarrow a^5+4\ge5a\)
\(a^4+1+1+1\ge4a\Rightarrow a^4+3\ge4a\)
\(\Rightarrow3\left(a^5+4\right)+4\left(a^4+3\right)\ge31a\)
\(\Rightarrow3\left(a^5+4\right)+4\left(a^4+3\right)+18a\ge49a\)
\(\Leftrightarrow\left(a^4+6\right)\left(3a+4\right)\ge49a\)
\(\Leftrightarrow\dfrac{a}{a^4+6}\le\dfrac{3a+4}{49}\)
Tương tự: \(\dfrac{b}{b^6+4}\le\dfrac{3b+4}{49}\) ; \(\dfrac{c}{c^4+6}\le\dfrac{3c+4}{49}\)
Cộng vế:
\(\dfrac{a}{a^4+6}+\dfrac{b}{b^4+6}+\dfrac{c}{c^4+6}\le\dfrac{3\left(a+b+c\right)+12}{49}=\dfrac{3}{7}\)
\(\dfrac{a}{a^4+6}+\dfrac{b}{b^4+6}+\dfrac{c}{c^4+6}\le\dfrac{3}{7}\left(1\right)\)
\(\Leftrightarrow\left(\dfrac{a}{a^4+6}-\dfrac{1}{7}\right)+\left(\dfrac{b}{b^4+6}-\dfrac{1}{7}\right)+\left(\dfrac{c}{c^4+6}-\dfrac{1}{7}\right)\le0\)
\(\Leftrightarrow\dfrac{a^4-7a+6}{a^4+6}+\dfrac{b^4-7b+6}{b^4+6}+\dfrac{c^4-7c+6}{c^4+6}\ge0\)
\(\Leftrightarrow\left(a-1\right).\dfrac{\left(a^3+a^2+a+1\right)}{a^4+6}+\left(b-1\right)\dfrac{\left(b^3+b^2+b+1\right)}{b^4+6}+\left(c-1\right)\dfrac{\left(c^3+c^2+c+1\right)}{c^4+6}\ge0\)
Với \(1\ge a\ge b\ge c>0\) ta có :
\(\left(a^3+a^2+a+1\right)\left(b^4+6\right)-\left(b^3+b^2+b+1\right)\left(a^4+6\right)\)
\(=\left(a^2+1\right)\left(a+1\right)\left(b^4+6\right)-\left(b^2+1\right)\left(b+1\right)\left(a^4+6\right)\ge0\)
\(\Rightarrow\dfrac{\left(a^3+a^2+a+1\right)}{a^4+6}\ge\dfrac{\left(b^3+b^2+b+1\right)}{b^4+6}\)
Giả sử \(a\ge b\ge c>0\Rightarrow\left(a-1\right)\ge\left(b-1\right)\ge\left(c-1\right)\) và \(\dfrac{\left(a^3+a^2+a+1\right)}{a^4+6}\ge\dfrac{\left(b^3+b^2+b+1\right)}{b^4+6}\ge\dfrac{\left(c^3+c^2+c+1\right)}{c^4+6}\)
\(\Rightarrow\) Áp dụng Bất đẳng thức Chebyshev, ta có :
\(\left(a-1+b-1+c-1\right)\left[\dfrac{\left(a^3+a^2+a+1\right)}{a^4+6}+\dfrac{\left(b^3+b^2+b+1\right)}{b^4+6}+\dfrac{\left(c^3+c^2+c+1\right)}{c^4+6}\right]=0\)
Vậy \(\left(1\right)\) đã được chứng minh
