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Let $a,b,c>0$ and $a+b+c=3$. Prove that: $P=\frac{{a}^{2}}{\sqrt{{b}^{3}+1}}+ \frac { { b }^{ 2 } }{ \sqrt{ { c }^{ 3 }+1 } }+\frac{{c}^{2}}{\sqrt{{a}^{3}+1}}\geq \frac{3\sqrt{2}}{2}$

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    Ok I have just clicking on the check box outline of all my questions :D thanks you so much2012-09-27

2 Answers 2

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This is going to be long so let's start

First, divide both sides by $\sqrt{2}$ and notice that by AM-GM $\sqrt{2(a^3+1)}\leq \frac{a^3+3}{2}.$ Similarly we proceed for the other variables and we find that $P\geq 2\left(\frac{a^2}{b^3+3}+\frac{b^2}{c^3+3}+\frac{c^2}{a^3+3}\right)=2\left(\frac{a^4}{a^2b^3+3a^2}+\frac{b^4}{b^2c^3+3b^2}+\frac{c^4}{c^2a^3+3a^2}\right).\tag{1}$

It is therefore sufficient to prove $\left(\frac{a^4}{a^2b^3+3a^2}+\frac{b^4}{b^2c^3+3b^2}+\frac{c^4}{c^2a^3+3a^2}\right)\geq\frac34.\tag{2}$

Use Cauchy Schwarz on the LHS of $(2)$ to find that $\left(\frac{a^4}{a^2b^3+3a^2}+\frac{b^4}{b^2c^3+3b^2}+\frac{c^4}{c^2a^3+3a^2}\right)\geq\frac{(a^2+b^2+c^2)^2}{a^2b^3+b^2c^3+c^2a^3+3(a^2+b^2+c^2)}.\tag{3}$

Again, it is sufficient to prove

$4(a^2+b^2+c^2)^2\geq 3(a^2b^3+b^2c^3+c^2a^3)+9(a^2+b^2+c^2).\tag{4}$

Now, since we may assume without loss of generality that $(b-a)(b-c)\leq 0$, the following is true

$\begin{split}a^{3}c^{2}+b^{3}a^{2}+c^{3}b^{2}&=a^{3}b^{2}+bc^{2}a^{2}+c^{3}b^{2}+a^{2}(b^{2}-c^{2})(b-a)\\&\leq a^{3}b^{2}+bc^{2}a^{2}+c^{3}b^{2}= ba^{2}\left(ab+\frac{1}{2}c^{2}\right)+bc^{2}\left(bc+\frac{1}{2}a^{2}\right)\\&\leq\frac{ba^{2}(a^{2}+b^{2}+c^{2})}{2}+\frac{bc^{2}(a^{2}+b^{2}+c^{2})}{2}\\&=\frac{a^{2}+b^{2}+c^{2}}{2\sqrt{2}}\sqrt{2b^{2}(a^{2}+c^{2})(a^{2}+c^{2})}\\&\leq\frac{a^{2}+b^{2}+c^{2}}{2\sqrt{2}}\sqrt{\left(\frac{2(a^{2}+b^{2}+c^{2})}{3}\right)^{3}}=3\sqrt{\left(\frac{a^{2}+b^{2}+c^{2}}{3}\right)^{5}},\end{split}\tag{5}$ it suffices to prove

$4(a^{2}+b^{2}+c^{2})^{2}\geq 9\sqrt{\left(\frac{a^{2}+b^{2}+c^{2}}{3}\right)^{5}}+9(a^{2}+b^{2}+c^{2})\tag{6}.$

Set $w:=\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}\tag{7}.$ By AM-QM, it immediately follows that $w\geq \frac{a+b+c}{3}=1,$ moreover, since $(a+b+c)^2>a^2+b^2+c^2,$ we also have $\sqrt 3>w.$ Rewrite $(6)$ using the parameter $w$ to obtain that our claim is equivalent to prove that $4w^{4}\geq w^{5}+3w^{2}\Leftrightarrow w^{2}(w-1)(3+3w-w^{2})\geq 0\tag{8}.$ This last assertion follows using our bounds on $w$, and therefore the proof is complete.

The equality case occurs when $a=b=c=1.$

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By AM-GM, C-S and Rearrangement we obtain: $\sum_{cyc}\frac{a^2}{\sqrt{b^3+1}}=\sum_{cyc}\frac{2\sqrt2a^2}{2\sqrt{2(b^2-b+1)(b+1)}}\geq\sum_{cyc}\frac{2\sqrt2a^2}{2(b^2-b+1)+(b+1)}=$ $=\sum_{cyc}\frac{2\sqrt2a^2}{2b^2-b+3}=\sum_{cyc}\frac{2\sqrt2a^4}{2a^2b^2-a^2b+3a^2}\geq\frac{2\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2b^2-a^2b+3a^2)}=$ $=\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(6a^2b^2-3a^2b+9a^2)}=\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(6a^2b^2-(a+b+c)a^2b+(a+b+c)^2a^2)}=$ $=\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(6a^2b^2-a^3b-a^2b^2-a^2bc+a^4+2a^2b^2+2a^3b+2a^3c+2a^2bc)}=$ $=\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+a^3b+2a^3c+7a^2b^2+a^2bc)}\geq\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^4+a^3b+a^3c+7a^2b^2+a^2bc)}.$ Thus, it remains to prove that $\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^4+a^3b+a^3c+7a^2b^2+a^2bc)}\geq\frac{3\sqrt2}{2}$ or $4(a^2+b^2+c^2)^2\geq\sum\limits_{cyc}(2a^4+a^3b+a^3c+7a^2b^2+a^2bc)$ or $\sum_{cyc}(2a^4-a^3b-a^3c+a^2b^2-a^2bc)\geq0$ or $\sum_{cyc}(a-b)^2(a^2+ab+b^2)+\frac{1}{2}\sum_{cyc}c^2(a-b)^2\geq0.$ Done!