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solakis1
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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
$\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\geq\frac{27}{2(a+b+c)^2}$
where a,b,c are positives
[sp] you may use the CAUCHY- SCHWARZ inequality[/sp]solakis said: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
lfdahl said: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 said:[sp]Should not the inequality be the other way round,i.e\[A^2 = \left ( \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \right )^2 \leq \left ( \frac{1}{a^2}+\frac{1}{b^2} +\frac{1}{c^2}\right )\left ( \frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2} \right )\][/sp]
[sp]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.[/sp]solakis said: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
The inequality means that the expression on the left side, which is the reciprocal of the sum of three products, is greater than or equal to the expression on the right side, which is the reciprocal of two times the square of the sum of three variables.
This inequality is important because it provides a way to compare the sum of three products to the square of the sum of three variables. It can also be used in various mathematical proofs and applications.
This inequality can be proven using various methods, such as algebraic manipulation, substitution, or induction. It can also be proven using geometric interpretations or calculus techniques.
This inequality has various applications in mathematics, including in number theory, algebra, and geometry. It can also be used in physics and engineering to solve problems involving sums of products.
Yes, there are exceptions to this inequality. For example, if any of the variables a, b, or c is equal to 0, then the inequality becomes undefined. Also, if the variables are negative, the inequality may not hold. Therefore, it is important to consider the domain of the variables when using this inequality.