<s>[INST] Prove that, if $a,b,c$ are positive real numbers $\frac {a} {2a+b+c}+ \frac {b} {2b+a+c}+\frac {c} {2c+a+b}<\frac{19}{25}$ 
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?
 [/INST] 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:
$$\sum_{cyc}\frac{a}{2a+b+c}\leq\frac{1}{(1+3)^2}
\sum_{cyc}a\left(\frac{1^2}{a}+\frac{3^3}{a+b+c}\right)=$$
$$=\frac{1}{16}\sum_{cyc}\left(1+\frac{9a}{a+b+c}\right)=\frac{3}{16}+\frac{9}{16}=\frac{3}{4}.$$
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