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Exercise 1: Let $a, b, c\ge 0$ satisfying $ab+bc+ca>0$. Find the minimum value of this expression: $P=\frac{1}{\sqrt{a(b+c)+2c^2}}+\frac{1}{\sqrt{b(a+4c)}}+2\sqrt{a+2b+4}+4\sqrt{c+1}$

Exercise 2: Let $a,b,c\ge 0$ satisfying $a^2+b^2+c^2=3$. Find Min of this expression? $\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}$

I think the Minimum of this case is $\frac{3}{2}$. I tried to solve by Cauchy-Schwarz but I can't prove that: $a^3+b^3+c^3+a^3b^2+b^3c^2+c^3a^2\le 6$ with $a^2+b^2+c^2=3$ ... Is this inequality right? Help me! Thanks!

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Lagrange multiplier? – Alex Dec 8 '13 at 18:15
You can show concretely for me? I need a simple solution. But if there is no other way, I'll use your way . Thanks – abcdxyz Dec 8 '13 at 18:25 Try following these examples. What do you get? – Alex Dec 8 '13 at 18:26
The inequality $a^3+b^3+c^3+a^3b^2+b^3c^2+c^3a^2\le 6$ with $a^2+b^2+c^2=3$ fails for $a=b=\sqrt{3/2}$ and $c=0$. – Alex Ravsky Dec 9 '13 at 4:15
Thanks .... I think the condition $a^2+b^2+c^2=3$, that my friend asked be failed ... If $a+b+c=3$,I can prove that by a reverse AM-GM inequality. P/s: Who can help me the Ex.1..? We can use $\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}$, and then use AM-GM to find Min? I'm not sure but i think so... – abcdxyz Dec 9 '13 at 5:50

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