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

x = 3 nha

điều kiện: \(x;y;z\ge\frac{1}{4}\)

cộng vế theo vế, ta được:

\(2x+2y+2z=\sqrt{4z-1}+\sqrt{4x-1}+\sqrt{4y-1}\)

\(\left(2x-\sqrt{4x-1}\right)+\left(2y-\sqrt{4y-1}\right)+\left(2z-\sqrt{4z-1}\right)=0\)

\(2\left(\sqrt{4x-1}-1\right)^2+2\left(\sqrt{4y-1}-1\right)^2+2\left(\sqrt{4z-1}-1\right)^2=0\)

ta có: \(VT\ge0\)với mọi x;y;z

dấu "=" xảy ra khi: \(\hept{\begin{cases}\sqrt{4x-1}=1\\\sqrt{4y-1}=1\\\sqrt{4z-1}=1\end{cases}}\), giải ra ta được: \(x=y=z=\frac{1}{2}\)(tmđk)

vậy...