Giúp mình!!
Cho a,b,c>0 thỏa mãn $\sqrt{a^{2}+b^{2}}+\sqrt{b^{2}+c^{2}}+\sqrt{c^{2}+a^{2}}$ $=$ $\sqrt{n}$
Chứng minh rằng $\frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b} \geq \frac{1}{2}.\sqrt{\frac{n}{2}}$
Từ điều kiện suy ra $\sqrt{n}=\sum_{cyc}\sqrt{a^2+b^2}\leq \sqrt{6(a^2+b^2+c^2)}$, hay $a^2+b^2+c^2\geq \frac{n}{6}$.
Từ đó có $\sum_{cyc}\frac{a^2}{b+c}\geq \sum_{cyc}\frac{a^2}{\sqrt{2(b^2+c^2)}} $ (theo bất đẳng thức C - S)
                                 $\geq \frac{1}{3}\sum_{cyc}a^2\sum_{cyc} \frac{1}{\sqrt{2(b^2+c^2)}}$ (theo bất đẳng thức Chebyshev)
                                 $\geq \frac{1}{3}\sum_{cyc} a^2\frac{9}{\sum_{cyc} \sqrt{2(b^2+c^2)}}$ (theo bất đẳng thức C - S)
                                 $\geq\frac{1}{3}.\frac{n}{6}.\frac{9}{\sqrt{2n}}$
                                 $\geq \frac{\sqrt{2n}}{4}$.
Đẳng thức xảy ra khi $a=b=c=\frac{\sqrt{2n}}{6}$.

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