prove that $\frac{1}{a(b + c)^{3}} + \frac{1}{b(c + a)^{3}} + \frac{1}{c(a + b)^{3}} \ge \frac{27}{8}$ given $a^{2} + b^{2} + c^{2} = 1$

Question

\begin{align} \text{Given } a, b, c > 0 \text{ with } a^{2} + b^{2} + c^{2} = 1, \text{ prove that} \\ \frac{1}{a(b + c)^{3}} + \frac{1}{b(c + a)^{3}} + \frac{1}{c(a + b)^{3}} \ge \frac{27}{8}. \end{align}

what i do so far

$$[(a + b) + (b + c) + (c + a)]\left[\frac{1}{a(b + c)^3} + \frac{1}{b(c + a)^3} + \frac{1}{c(a + b)^3}\right] \ge \left(\frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}}\right)^2.$$

$$\frac{1}{\sqrt{a}} + \frac{1}{\sqrt{b}} + \frac{1}{\sqrt{c}} \ge 3 \sqrt[3]{\frac{1}{\sqrt{a} \sqrt{b} \sqrt{c}}} = 3\sqrt[6]{\frac{1}{abc}}.$$

$$[(a + b) + (b + c) + (c + a)]\left[\frac{1}{a(b + c)^3} + \frac{1}{b(c + a)^3} + \frac{1}{c(a + b)^3}\right] \ge \left(3\sqrt[6]{\frac{1}{abc}}\right)^2.$$

$$2(a + b + c)\left[\frac{1}{a(b + c)^3} + \frac{1}{b(c + a)^3} + \frac{1}{c(a + b)^3}\right] \ge 27 \sqrt[3]{\frac{1}{abc}}.$$

$$2(a + b + c)\left[\frac{1}{a(b + c)^3} + \frac{1}{b(c + a)^3} + \frac{1}{c(a + b)^3}\right] \ge 27 \sqrt[3]{\frac{1}{abc}}$$ $$2(a + b + c)\left[\frac{1}{a(1 - a)^3} + \frac{1}{b(1 - b)^3} + \frac{1}{c(1 - c)^3}\right] \ge 27 \sqrt[3]{\frac{1}{abc}}.$$

I Apply AM-GM Inequality to $(a + b + c)$

$ 2(a + b + c) \ge 6\sqrt[3]{abc}.$

$$6\sqrt[3]{abc}\left[\frac{1}{a(1 - a)^3} + \frac{1}{b(1 - b)^3} + \frac{1}{c(1 - c)^3}\right] \ge 27 \sqrt[3]{\frac{1}{abc}}.$$

and sorry, because I am new on MSE so if there are any mistakes on how to ask a qestion

Answer 1

The idea is to homogenize and use Holder’s inequality

Consider:

The LHS of the problem is equal to

(Sum_{cyc} 1/(a(b+c)^3)) (Sum_{cyc} a^2)^2>= [Sum_{cyc} [((a^2)^2)(1/(a(b+c)^3))]^(1/3)]^3

Which simplifies to

(a/(b+c)+b/(a+c)+c/(a+b))^3 >= 27/8 iff a/(b+c)+b/(a+c)+c/(a+b) >= 3/2

Which is the famous Nesbitt’s inequality.

Note: A motivation for multiplying (a^2+b^2+c^2)^2 is because it directly homogenizes the inequality.

Answer 2

Since $f(x)=\frac{1}{\sqrt{x}\left(\frac{1-x}{2}\right)^{\frac{3}{2}}}$ is a convex function (because $f''(x)=\frac{3}{8\sqrt{2x^5(1-x)}}>0$),

by P-M and Jensen we obtain: $$\sum_{cyc}\tfrac{1}{a(b+c)^3}=\tfrac{1}{8}\sum_{cyc}\tfrac{1}{a\left(\frac{b+c}{2}\right)^3}\geq\tfrac{1}{8}\sum_{cyc}\tfrac{1}{a\left(\frac{b^2+c^2}{2}\right)^\frac{3}{2}}=\frac{1}{8}\sum_{cyc}\tfrac{1}{a\left(\frac{1-a^2}{2}\right)^{\frac{3}{2}}}\geq\frac{1}{8}\cdot\tfrac{3}{\sqrt{\frac{1}{3}}\left(\frac{1-\frac{1}{3}}{2}\right)^{\frac{3}{2}}}=\tfrac{27}{8}.$$

Answer 3

Another way.

By AM-GM and C-S we obtain: $$\sum_{cyc}\frac{1}{a(b+c)^3}\geq\frac{3}{\sqrt[3]{abc}(a+b)(a+c)(b+c)}\geq\frac{3}{\frac{a+b+c}{3}\cdot\left(\frac{\sum\limits_{cyc}(a+b)}{3}\right)^3}=$$ $$=\frac{3}{8\left(\frac{a+b+c}{3}\right)^4}=\frac{3}{8\left(\frac{a^2+b^2+c^2}{3}\right)^2}=\frac{27}{8}.$$

Answer 4

Lemma $1$ (modified Nesbitt): $\sum_{\text{cyc}} \frac{b+c}{a} \ge 6$. Proof: By Hölder, $\left(\sum_{\text{cyc}} \frac{b+c}{a}\right)\cdot\left(\sum_{\text{cyc}} a(b+c)\right) \ge (2(a+b+c))^2 \iff \sum_{\text{cyc}} \frac{b+c}{a} \ge \frac{4(a+b+c)^2}{2(ab+ac+bc)} \ge \frac{4(3\cdot(ab+ac+bc))}{2(ab+ac+bc)}= 6. \hspace{1cm} \square$

We now have:

\begin{equation} \left(\sum_{\text{cyc}} \frac{1}{a(b+c)^3}\right)\cdot\left(\sum_{\text{cyc}}\frac{b+c}{a}\right) \overset{\mathrm{Hölder}}{\ge} \left(\sum_{\text{cyc}} \frac{1}{a(b+c)}\right)^2 \overset{\mathrm{Lemma}}{\ge}\frac{1}{6}\left(\sum_{\text{cyc}} \frac{1}{a(b+c)}\right)^2 \end{equation}

And it hence suffices to prove

\begin{equation} \left(\sum_{\text{cyc}} \frac{1}{a(b+c)}\right)^2 \ge 6\cdot\frac{27}{8} = \frac{81}{4} \iff \sum_{\text{cyc}} \frac{1}{a(b+c)} \ge \frac{9}{2} \hspace{3mm}\overset{\mathrm{a^2+b^2+c^2=1}}{\iff} \sum_{\text{cyc}} \frac{a^2+b^2+c^2}{a(b+c)} \ge \frac{9}{2}. \\ \sum_{\text{cyc}} \frac{a^2+b^2+c^2}{a(b+c)} \ge \sum_{\text{cyc}} \frac{ab+ac+bc}{ab+ac} = 3 + \sum_{\text{cyc}} \frac{bc}{ab+ac} \overset{\mathrm{Nesbitt}}{\ge} 3 + \frac{3}{2} = \frac{9}{2}. \hspace{5mm} \square \end{equation}

Done

Answer 5

The function is minimized when all values are equal (consider maximizing $a(b+c)^3+..$) so $a = \frac{1}{\sqrt{3}}$ at minimum.

Evaluating the function at the minimum gives: $$ \frac{3\sqrt{3}}{(\frac{2}{\sqrt{3}})^3} = \frac{3^3}{8} = \frac{27}{8}. $$