Chủ đề này có 1 trả lời

[doctor lee]

cho a,b,c>0 tm $a^{2}+b^{2}+c^{2}=3$

tìm max $P= \frac{a}{\sqrt{2-bc}}+\frac{b}{\sqrt{2-ac}}+\frac{c}{\sqrt{2-ab}}$

[melodias2002]

$\sum \frac{a}{\sqrt{2-bc}} \leq \sum \frac{a}{\sqrt{2-\frac{b^2+c^2}{2}}} =\sum \frac{a}{\sqrt{2-\frac{3-a^2}{2}}}=\sum \frac{\sqrt{2}a}{\sqrt{a^2+1}}\leq\sum \frac{\sqrt{2}a}{\sqrt{2a}}=\sum \sqrt{a}$

Mà $\sum \sqrt{a}\leq\sqrt{3(a+b+c)}\leq\sqrt{3\sqrt{3(a^2+b^2+c^2)}}=3$

Suy ra  $\sum \frac{a}{\sqrt{2-bc}} \leq 3$

Dấu "$=$" xảy ra $\Leftrightarrow a=b=c=1$