If $a,b,c > 0$, then?

Question

If $a,b,c > 0$, then: $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}\ ?$$

Can I get a step by step solution please ?

Answer 1

HINT: Setting $$b+c=x,a+c=y,a+b=z$$ then we get $$a=\frac{-x+y+z}{2}$$ $$b=\frac{x-y+z}{2}$$ $$c=\frac{x+y-z}{2}$$ then we get $$\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}\geq 6$$ which is true.

Answer 2

Set $\;A=b+c$, $\;B=c+a$, $\;C=a+b$, $\;S=A+B+C$. Note that $S=2(a+b+c)$. We can rewrite the l.h.s. expression as $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{\dfrac S2-A}{A}+\frac{\dfrac S2-B}{B}+\frac{\dfrac S2-C}{C}=\frac S2\Bigl(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\Bigr)-3$$ If we denote $H$ the harmonic mean of $A$, $B$, $C$ and $\mathcal A$ their arithmetic mean, this may be re-written as $$\frac {3\mathcal A}2\cdot\frac 3H\ge \frac92$$ by the arithmetic-harmonic mean inequality, so that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac 92-3=\frac32.$$

Answer 3

As always these have to do with setting up the $AM-GM$ thereom that states $\frac {(\sum_{i=1}^n a_i)}{n}\ge \sqrt[n]{\prod_{i=1}^n a_i}$. The hard part is the exact set up.

I personally don't like fractions of the form $\frac {a}{b+c}$ and would prefer $\frac {x + y}{z}$ where $z = b+c$ and $a = ....$ something.

Let $x = a+b; y = b+c; z = a+c$ then $2a = x+z - y$ etc.

$\frac {a}{b+ c} + \frac b{a+c} + \frac c{a+b} = \frac 12(\frac{x+z-y}{y} + \frac {x+y - z}{z} + \frac {y+z - x}{x})=$

$\frac 12(\frac xy + \frac zy + \frac xz + \frac yz + \frac yx + \frac zx - 3)$

Bear with me: If we let $D=\frac xy$ and $E=\frac xz$ and $F=\frac yz$ then we have

$\frac 12(\frac xy + \frac zy + \frac xz + \frac yz + \frac yx + \frac zx - 3)=$

$\frac 12(D + \frac 1D + E + \frac 1E + F + \frac 1F - 3)$.

You may have seen the result that for $x > 0$ that $x + \frac 1x \ge 2$. That is a direct application of AM-GM: $\frac{x + \frac 1x}2 \ge \sqrt{x*\frac 1x} = \sqrt 1 =1$. (There are other ways of proving it: I'm fond of $(x-1)^2 \ge 0$ so $x^2 +1 \ge 2x$ so $x + \frac 1x \ge 2$[$*$].)

But anyhow... that means:

$\frac 12(D + \frac 1D + E + \frac 1E + F + \frac 1F - 3)\ge \frac 12(2+2+2 -3)=\frac 32$

[$*$] This raises the question, how do we prove AM-GM in the first place. Well, for $2$ values $a> 0; b>0$, the $(\sqrt{a} - \sqrt{b})^2 \ge 0$ so $a + b - 2\sqrt{a}\sqrt{b} > 0$ so $\frac {a+b}2 \ge \sqrt{ab}$. For more values, it follows by induction and the binomial theorem so... those two proofs that $x + \frac 1x \ge 2$ are really the same thing.