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inequality

solakis

Active member
Dec 9, 2012
304
Prove:

$\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\geq\frac{27}{2(a+b+c)^2}$

where a,b,c are positives
 

solakis

Active member
Dec 9, 2012
304
Prove:

$\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\geq\frac{27}{2(a+b+c)^2}$

where a,b,c are positives

you may use the CAUCHY- SCHWARZ inequality
 

lfdahl

Well-known member
Nov 26, 2013
719
My attempt:


Applying the Cauchy-Schwarz inequality yields:

\[A^2 = \left ( \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \right )^2 \geq \left ( \frac{1}{a^2}+\frac{1}{b^2} +\frac{1}{c^2}\right )\left ( \frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2} \right )\]

From the fact, that the root-mean-square is always greater than or equal to the harmonic mean, I get:
\[\frac{1}{a^2}+\frac{1}{b^2} +\frac{1}{c^2} \geq \frac{3^3}{(a+b+c)^2}\]
- and
\[\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2} +\frac{1}{(a+c)^2} \geq \frac{3^3}{2^2(a+b+c)^2}\]

Thus,

\[A^2 \geq \frac{3^6}{2^2(a+b+c)^4}\]

- or

\[A = \frac{1}{b(a+b)} +\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \geq \frac{27}{2(a+b+c)^2}\]

 

solakis

Active member
Dec 9, 2012
304
My attempt:


Applying the Cauchy-Schwarz inequality yields:

\[A^2 = \left ( \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \right )^2 \geq \left ( \frac{1}{a^2}+\frac{1}{b^2} +\frac{1}{c^2}\right )\left ( \frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2} \right )\]

From the fact, that the root-mean-square is always greater than or equal to the harmonic mean, I get:
\[\frac{1}{a^2}+\frac{1}{b^2} +\frac{1}{c^2} \geq \frac{3^3}{(a+b+c)^2}\]
- and
\[\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2} +\frac{1}{(a+c)^2} \geq \frac{3^3}{2^2(a+b+c)^2}\]



Thus,

\[A^2 \geq \frac{3^6}{2^2(a+b+c)^4}\]

- or

\[A = \frac{1}{b(a+b)} +\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \geq \frac{27}{2(a+b+c)^2}\]

Should not the inequality be the other way round,i.e


\[A^2 = \left ( \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \right )^2 \leq \left ( \frac{1}{a^2}+\frac{1}{b^2} +\frac{1}{c^2}\right )\left ( \frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2} \right )\]
 

lfdahl

Well-known member
Nov 26, 2013
719
Should not the inequality be the other way round,i.e


\[A^2 = \left ( \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \right )^2 \leq \left ( \frac{1}{a^2}+\frac{1}{b^2} +\frac{1}{c^2}\right )\left ( \frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2} \right )\]

You´re absolutely right! It´s my mistake. Thankyou for pointing out the error.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,684
Prove:

$\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\geq\frac{27}{2(a+b+c)^2}$

where a,b,c are positives
Following the hint, apply the Cauchy-Schwarz inequality to the vectors $\bigl(\sqrt{a+b},\sqrt{b+c},\sqrt{c+a}\bigr)$ and $\bigl(\frac1{\sqrt{b(a+b)}},\frac1{\sqrt{c(b+c)}},\frac1{\sqrt{a(c+a)}}\bigr)$. That gives $$\left(\frac1{\sqrt b} + \dfrac1{\sqrt c} + \dfrac1{\sqrt a}\right)^2 \leqslant 2(a+b+c)\left(\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\right).$$ The required result will follow if we can show that $$27 \leqslant (a+b+c)\left(\frac1{\sqrt a} + \dfrac1{\sqrt b} + \dfrac1{\sqrt c}\right)^2.$$ To do that, I need to use Hölder's inequality, which (applied to vectors in 3-dimensional space, with all coordinates positive) says that if $\mathbf{x} = (x_1,x_2,x_3)$, $\mathbf{y} = (y_1,y_2,y_3)$ and $\frac1p + \frac1q = 1$, then $$x_1y_1 + x_2y_2 + x_3y_3 \leqslant (x_1^p+x_2^p+x_3^p)^{1/p} (y_1^q+y_2^q+y_3^q)^{1/q}.$$ With $p=3$, $q=3/2$, $\mathbf{x} = (a^{1/3},b^{1/3},c^{1/3})$ and $\mathbf{y} = (a^{-1/3},b^{-1/3},c^{-1/3})$, that becomes $$3 \leqslant (a+b+c)^{1/3}\left(\frac1{\sqrt a} + \dfrac1{\sqrt b} + \dfrac1{\sqrt c}\right)^{2/3}.$$ After cubing both sides, that gives the result.
 

steep

Member
Dec 17, 2018
51
a different take would be to look at this in terms of convexity.

$\mu :=\frac{a+b+c}{3}$
the inequality is equivalent to proving
$\frac{1}{2\mu^2} = \frac{1}{\mu(\mu + \mu)} \leq \frac{1}{3}\Big(\frac{1}{b(a+b)} + \frac{1}{c(b+c)} + \frac{1}{a(c+a)}\Big)$


if we consider the function $f: \mathbb R^3 \mapsto \mathbb R$ given by
$f\big(\mathbf x\big) = \frac{1}{x_2(x_1 + x_2)}$

we can examine its Hessian
$\mathbf H =\left[\begin{matrix}\frac{2}{x_{2} \left(x_{1} + x_{2}\right)^{3}} & \frac{2}{x_{2} \left(x_{1} + x_{2}\right)^{3}} + \frac{1}{x_{2}^{2} \left(x_{1} + x_{2}\right)^{2}} & 0\\\frac{2}{x_{2} \left(x_{1} + x_{2}\right)^{3}} + \frac{1}{x_{2}^{2} \left(x_{1} + x_{2}\right)^{2}} & \frac{2}{x_{2} \left(x_{1} + x_{2}\right)^{3}} + \frac{2}{x_{2}^{2} \left(x_{1} + x_{2}\right)^{2}} + \frac{2}{x_{2}^{3} \left(x_{1} + x_{2}\right)} & 0\\0 & 0 & 0\end{matrix}\right]$

and e.g. apply Sylvester's Determinant Criterion to confirm that $f$ is convex so long as each $x_i \gt 0$, i.e.

$\det\big(\mathbf H_{1:1}\big) = \frac{2}{x_{2} (x_{1} + x_{2})^{3}} \gt 0$ since each component is positive

$\det\big(\mathbf H_{2:2}\big) = \frac{3}{x_2^4(x_1 + x_2)^4}$
https://www.wolframalpha.com/input/?i=hessian+of+1/(x_2*(x_1+x_2))+

$\det\big(\mathbf H_{3:3}\big) = \det\big(\mathbf H\big) = 0$
because there is a column of all zeros

now, selecting:
$\mathbf x := \left[\begin{matrix} a \\ b \\ c\end{matrix}\right] $
and using cyclic permutation matrix

$\mathbf P = \left[\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}\right] $

noting that
$\mathbf P^0 + \mathbf P^1 + \mathbf P^2 = \left[\begin{matrix}1 & 1 & 1\\1 & 1 & 1\\1 & 1 & 1\end{matrix}\right]$

we get
$\frac{1}{\mu(\mu + \mu)}= f\Big(\frac{1}{3}\big(\mathbf P^0 + \mathbf P^1+\mathbf P^2\big)\mathbf x\Big) = f\Big(\frac{1}{3}\big(\mathbf P^0\mathbf x + \mathbf P^1\mathbf x +\mathbf P^2\mathbf x\big)\Big)$
$ \leq \frac{1}{3}\Big(f\big(\mathbf P^0\mathbf x\big) + f\big(\mathbf P^1\mathbf x\big)+ f\big(\mathbf P^2\mathbf x\big) \Big) = \frac{1}{3}\Big(\frac{1}{b(a+b)}+ \frac{1}{a(c+a)} + \frac{1}{c(b+c)}\Big) = \frac{1}{3}\Big(\frac{1}{b(a+b)} + \frac{1}{c(b+c)} + \frac{1}{a(c+a)}\Big)$

by Jensen's Inequality


technical items:
i.) the components of $\mathbf H$ are rational functions so I take for granted that they vary continuously with $\mathbf x$
ii.) Sylvester's Determinant criterion might seem 'wrong' here since it technically applies when all determinants are positive and our final one is zero-- but since the first 2 leading minors are positive, it applies to that 2x2 principal submatrix and implies that is positive definite e.g. it implies the following Cholesky factorization with blocked structure showing positive semi-definiteness for $\mathbf H$

$\mathbf H =\left[\begin{matrix}\frac{2}{x_{2} \left(x_{1} + x_{2}\right)^{3}} & \frac{2}{x_{2} \left(x_{1} + x_{2}\right)^{3}} + \frac{1}{x_{2}^{2} \left(x_{1} + x_{2}\right)^{2}} & 0\\\frac{2}{x_{2} \left(x_{1} + x_{2}\right)^{3}} + \frac{1}{x_{2}^{2} \left(x_{1} + x_{2}\right)^{2}} & \frac{2}{x_{2} \left(x_{1} + x_{2}\right)^{3}} + \frac{2}{x_{2}^{2} \left(x_{1} + x_{2}\right)^{2}} + \frac{2}{x_{2}^{3} \left(x_{1} + x_{2}\right)} & 0\\0 & 0 & 0\end{matrix}\right]= \left[\begin{matrix} \mathbf {LL}^T & 0 \\ 0 & 0 \end{matrix}\right] = \left[\begin{matrix} \mathbf {L} & 0 \\ 0 & 0 \end{matrix}\right]\left[\begin{matrix} \mathbf {L} & 0 \\ 0 & 0 \end{matrix}\right]^T \succeq 0$
 

solakis

Active member
Dec 9, 2012
304
$\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\geq\frac{27}{2(a+b+c)^2}$.........................................................................................(1)

or
$2(a+b+c)(\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)})\geq\frac{27}{(a+b+c)}$

But :

$2(a+b+c)(\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)})$ =

=$[(\sqrt(a+b))^2+(\sqrt (b+c))^2+(\sqrt (c+a))^2][(\frac{1}{\sqrt b(a+b)})^2+(\frac{1}{\sqrt c(b+c)})^2+(\frac{1}{\sqrt a(c+a)})^2]$

Which according to the B-C-S inequality is greater or equal to:

$(\frac{1}{\sqrt b} +\frac{1}{\sqrt c}+\frac{1}{\sqrt a})^2$


Hence we have to prove:




$(\frac{1}{\sqrt b} +(\frac{1}{\sqrt c}+\frac{1}{\sqrt a})^2\geq \frac{27}{a+b+c)}$ to satisfy (1)

or:

$(a+b+c)(\frac{1}{\sqrt b} +\frac{1}{\sqrt c}+\frac{1}{\sqrt a})^2\geq 27$................................................................................(2)


But from the inequalities:

$(\sqrt a-\sqrt b)^2\geq 0$, $(\sqrt b-\sqrt c)^2\geq 0$, $\sqrt c-\sqrt a)^2)\geq 0$

we can get the inequality:

$(a+b+c)\geq\frac{(\sqrt a+\sqrt b+\sqrt c)^2}{3}$
and multiplying both sides of the above inequality by$ (\frac{1}{\sqrt b} +\frac{1}{\sqrt c}+\frac{1}{\sqrt a})^2$ we have:


$(a+b+c)(\frac{1}{\sqrt b} +\frac{1}{\sqrt c}+\frac{1}{\sqrt a})^2\geq\frac{(\sqrt a+\sqrt b+\sqrt c)^2}{3}(\frac{1}{\sqrt b} +\frac{1}{\sqrt c}+\frac{1}{\sqrt a})^2$



Hence to satisfy (2) and thus (1) we have to prove that:


$\frac{(\sqrt a+\sqrt b+\sqrt c)^2}{3}(\frac{1}{\sqrt b} +\frac{1}{\sqrt c}+\frac{1}{\sqrt a})^2\geq 27$

or


$(\sqrt a+\sqrt b+\sqrt c)(\frac{1}{\sqrt b} +\frac{1}{\sqrt c}+\frac{1}{\sqrt a})\geq 9$

Which is true by using again the B-C-S inequality where we put:

$x_1$=$a^\frac{1}{4}$ $x_2=\frac{1}{a^\frac{1}{4}}$

$y_1=b^\frac{1}{4}$ $y_2= \frac{1}{b^\frac{1}{4}}$

$z_1=c^\frac{1}{4}$ $z_2= \frac{1}{b^\frac{1}{4}}$