\(\left(\frac{1}{2\sqrt{2}}c-\sqrt{2}\right)^2\ge0\)
\(\Rightarrow\frac{1}{8}c^2-c+2\ge0\)
\(\Rightarrow\frac{1}{2\sqrt{2}}c\ge\sqrt{c-2}\)
\(\Rightarrow\frac{1}{2\sqrt{2}}\ge\frac{\sqrt{c-2}}{c}\)
tương tự \(\left(\frac{1}{2\sqrt{3}}a-\sqrt{3}\right)^2\ge0\Rightarrow\frac{1}{2\sqrt{3}}\ge\frac{\sqrt{a-3}}{a}\)
\(\left(\frac{1}{4}b-2\right)^2\ge0\Rightarrow\frac{1}{4}\ge\frac{\sqrt{b-4}}{b}\)
\(\Rightarrow P\le\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+\frac{1}{4}\)
Dấu "=" xảy ra <=>c=4;a=6;b=8