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