\(\left(a^3+b\right)\left(a+b^3\right)\ge\left(a^2+b^2\right)^2\Rightarrow\left\{{}\begin{matrix}\frac{1}{a^3+b}\le\frac{a+b^3}{\left(a^2+b^2\right)^2}\\\frac{1}{a+b^3}\le\frac{a^3+b}{\left(a^2+b^2\right)^2}\end{matrix}\right.\)
\(\Rightarrow P\le\left(a+b\right)\left(\frac{a+b+a^3+b^3}{\left(a^2+b^2\right)^2}\right)-\frac{1}{ab}=\frac{\left(a+b\right)^2\left(a^2+b^2-ab+1\right)}{\left(a^2+b^2\right)^2}-\frac{1}{ab}\)
\(P\le\frac{\left(a+b\right)^2\left(a^2+b^2+1-ab\right)}{\frac{\left(a+b\right)^2}{2}.\left(a^2+b^2\right)}-\frac{1}{ab}=2+\frac{2-2ab}{a^2+b^2}-\frac{1}{ab}\)
\(\Rightarrow P\le2+\frac{2-2ab}{2ab}-\frac{1}{ab}=1\)
Dấu "=" xảy ra khi \(a=b=1\)
\(12x^2+26xy+15y^2=4617\Leftrightarrow11\left(x+y\right)^2+\left(x+2y\right)^2=4617\)
\(11\left(x+y\right)^2⋮11\) và \(4617\) chia 11 dư 8
\(\Rightarrow\left(x+2y\right)^2\) chia 11 dư 8 (vô lý)
Vậy pt ko có nghiệm nguyên
