<s>[INST] How prove this inequality $\frac{a^2+bc}{3+a}+\frac{b^2+ca}{3+b}+\frac{c^2+ab}{3+c}\le \frac{7}{4}$ let $a,b,c\ge 0$,show that
$$\dfrac{a^2+bc}{2a+b+c}+\dfrac{b^2+ca}{2b+c+a}+\dfrac{c^2+ab}{2c+a+b}\le\dfrac{7}{12}(a+b+c)$$
My idea: Without loss of generality,we Assmue that
$$a+b+c=3$$
then we only prove this 
$$\dfrac{a^2+bc}{3+a}+\dfrac{b^2+ca}{3+b}+\dfrac{c^2+ab}{3+c}\le \dfrac{7}{4}$$
then I can't,Thank you very much
 [/INST] After full expanding we need to prove that
$$\sum_{cyc}(2a^4+3a^3b+3a^3c-10a^2b^2+66a^2bc)\geq0$$ or
$$2\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+5\sum_{cyc}ab(a-b)^2+64abc(a+b+c)\geq0,$$
which is true by Schur.
Done!
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