Ta có: \(P=\frac{25a^2}{\sqrt{2a^2+16ab+7b^2}}+\frac{25b^2}{\sqrt{2b^2+16bc+7c^2}}+\frac{c^2\left(3+a\right)}{a}\)\(=\frac{25a^2}{\sqrt{\left(2a+3b\right)^2-2\left(a-b\right)^2}}+\frac{25b^2}{\sqrt{\left(2b+3c\right)^2-2\left(b-c\right)^2}}+\frac{c^2\left(3+a\right)}{a}\)\(\ge\frac{25a^2}{2a+3b}+\frac{25b^2}{2b+3c}+\frac{c^2\left(3+a\right)}{a}\)
Áp dụng bất đẳng thức AM - GM, ta có: \(\frac{25a^2}{2a+3b}+\left(2a+3b\right)\ge2\sqrt{\frac{25a^2}{2a+3b}.\left(2a+3b\right)}=10a\Rightarrow\frac{25a^2}{2a+3b}\ge8a-3b\)(1)
\(\frac{25b^2}{2b+3c}+\left(2b+3c\right)\ge2\sqrt{\frac{25b^2}{2b+3c}.\left(2b+3c\right)}=10b\Rightarrow\frac{25b^2}{2b+3c}\ge8b-3c\)(2)
\(\frac{c^2\left(3+a\right)}{a}=\frac{3c^2}{a}+c^2=\left(\frac{3c^2}{a}+3a\right)+\left(c^2+1\right)-3a-1\)\(\ge2\sqrt{\frac{3c^2}{a}.3a}+2c-3a-1=8c-3a-1\)(3)
Cộng theo vế ba bất đẳng thức (1), (2), (3), ta được: \(\frac{25a^2}{2a+3b}+\frac{25b^2}{2b+3c}+\frac{c^2\left(3+a\right)}{a}\ge5\left(a+b+c\right)-1=14\)
Vậy \(P\ge14\)
Đẳng thức xảy ra khi a = b = c = 1
