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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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I don't know if you are aware of it, but you have a $0\%$ accept rate. –  Aang Sep 27 '12 at 6:41
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Expanding 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... –  Dennis Gulko Sep 27 '12 at 8:28
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@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? –  Dennis Gulko Sep 27 '12 at 8:46
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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. * –  draks ... Sep 27 '12 at 9:09
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Ok I have just clicking on the check box outline of all my questions :D thanks you so much –  LevanDokite Sep 27 '12 at 9:13

1 Answer 1

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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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