Let $a,b,c>0$ and $a+b+c=3$. Prove that: $$P=\frac{{a}^{2}}{\sqrt{{b}^{3}+1}}+ \frac { { b }^{ 2 } }{ \sqrt{ { c }^{ 3 }+1 } }+\frac{{c}^{2}}{\sqrt{{a}^{3}+1}}\geq \frac{3\sqrt{2}}{2}$$
Let $a,b,c>0$ and $a+b+c=3$. Find the minimum of $P$.
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0i 've repaired aldready – 2012-09-27
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2Expanding on Avatar's comment: If none of the answers you received so far on this site are satisfactory to you, maybe you should stop asking questions here... – 2012-09-27
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0I don't know the answer, what are you talking about? – 2012-09-27
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2@anmaylamgiau: You asked 7 questions on this site. You did not accept any of the answers that you got, which means that you are not satisfied with with any of them, is that correct? – 2012-09-27
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0"You did not accept any of the answers that you got" sorry I don't know the rules of this site – 2012-09-27
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1http://math.stackexchange.com/faq#howtoask: *When you have decided which answer is the most helpful to you, mark it as the accepted answer by clicking on the check box outline to the left of the answer. * – 2012-09-27
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2Ok I have just clicking on the check box outline of all my questions :D thanks you so much – 2012-09-27
2 Answers
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 $a=b=c=1.$
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!