Áp dụng giả thiết và bất đẳng thức AM - GM, ta có: \(VT=\frac{x}{\sqrt{x^2+xy+yz+zx}}+\frac{y}{\sqrt{y^2+xy+yz+zx}}+\frac{z}{\sqrt{z^2+xy+yz+zx}}\)\(=\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\frac{y}{\sqrt{\left(y+x\right)\left(y+z\right)}}+\frac{z}{\sqrt{\left(z+x\right)\left(z+y\right)}}\)\(=\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}+\sqrt{\frac{y}{x+y}.\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}.\frac{z}{y+z}}\)\(\le\frac{\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}+\frac{z}{y+z}}{2}=\frac{3}{2}\)

Đẳng thức xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)

Theo nguyên lí Dirichlet thì trong 3 số 2a - 1, 2b - 1, 2c - 1 tồn tại ít nhất 2 số cùng dấu

Giả sử đó là 2a - 1 và 2b - 1 thì \(\left(2a-1\right)\left(2b-1\right)\ge0\Leftrightarrow4ab-2a-2b+1\ge0\Leftrightarrow2ab\ge a+b-\frac{1}{2}\)\(\Leftrightarrow2abc\ge ac+bc-\frac{c}{2}\)

\(P=ab+bc+ca-abc=ab+bc+ca-2abc+abc\)\(\le ab+bc+ca-ac-bc+\frac{c}{2}+abc=ab+abc+\frac{c}{2}\)\(\le\frac{a^2+b^2}{2}+abc+\frac{c}{2}+\frac{c^2}{2}-\frac{c^2}{2}-\frac{1}{8}+\frac{1}{8}\)\(=\frac{a^2+b^2+c^2+2abc}{2}-\frac{1}{2}\left(c-\frac{1}{2}\right)^2+\frac{1}{8}\le\frac{1}{2}+\frac{1}{8}=\frac{5}{8}\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{2}\)

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