Prove that, if $a,b,c$ are positive real numbers $\frac {a} {2a+b+c}+ \frac {b} {2b+a+c}+\frac {c} {2c+a+b}<\frac{19}{25}$

Question

Prove that, if $a,b,c$ are positive real numbers:

$$\frac {a} {2a+b+c}+ \frac {b} {2b+a+c}+\frac {c} {2c+a+b}<\frac{19}{25}$$

Can the proof be written without the need for high mathematics?

Answer 1

The inequality is homogeneous, i.e. $(a,b,c)$ satisfies the inequality if and only if $(ta,tb,tc)$ $\forall t>0$ satisfies the inequality. So we can assume WLOG $a+b+c=1$.

$$\sum_{\text{cyc}}\frac{a}{2a+b+c}=\sum_{\text{cyc}}\frac{a}{a+1}=$$

$$=\sum_{\text{cyc}}\left(1-\frac{1}{a+1}\right)=3-\sum_{\text{cyc}}\frac{1}{a+1}$$

You can use Cauchy Schwarz (CS) inequality to prove that for real $a_i$, positive $b_i$:

$$\sum_{i=1}^n\frac{a_i^2}{b_i}\ge \frac{(\sum_{i=1}^n a_i)^2}{\sum_{i=1}^n b_i}$$

To prove it, multiply both sides by $\sum_{i=1}^n b_i$ and use CS.

$$3-\sum_{\text{cyc}}\frac{1}{a+1}\le 3-\frac{9}{4}=\frac{3}{4}<\frac{19}{25}$$

Answer 2

We'll prove a stronger inequality: $$\sum_{cyc}\frac{a}{2a+b+c}\leq\frac{3}{4}$$ or $$\sum_{cyc}\left(\frac{1}{4}-\frac{a}{2a+b+c}\right)\geq0$$ or $$\sum_{cyc}\frac{b+c-2a}{2a+b+c}\geq0$$ or $$\sum_{cyc}\frac{c-a-(a-b)}{2a+b+c}\geq0$$ or $$\sum_{cyc}\left(\frac{c-a}{2a+b+c}-\frac{a-b}{2a+b+c}\right)\geq0$$ or $$\sum_{cyc}\left(\frac{a-b}{2b+c+a}-\frac{a-b}{2a+b+c}\right)\geq0$$ or

$$\sum_{cyc}\frac{(a-b)^2}{(2a+b+c)(2b+a+c)}\geq0.$$ Done!

Also, we can use C-S: $$\sum_{cyc}\frac{a}{2a+b+c}\leq\frac{1}{(1+3)^2} \sum_{cyc}a\left(\frac{1^2}{a}+\frac{3^3}{a+b+c}\right)=$$ $$=\frac{1}{16}\sum_{cyc}\left(1+\frac{9a}{a+b+c}\right)=\frac{3}{16}+\frac{9}{16}=\frac{3}{4}.$$