If $a^2+b^2+c^2=3,$ then prove $\sqrt{\frac{a+b}{4a+3}}+\sqrt{\frac{b+c}{4b+3}}+\sqrt{\frac{c+a}{4c+3}}\le 3\sqrt{\frac{2}{7}}.$

Question

Question. Given non-negative real numbers $a,b,c$ such that $a^2+b^2+c^2=3.$ Prove that $$\sqrt{\frac{a+b}{4a+3}}+\sqrt{\frac{b+c}{4b+3}}+\sqrt{\frac{c+a}{4c+3}}\le 3\sqrt{\frac{2}{7}}.$$ I tried to use Cauchy-Schwarz and we need to prove $$\frac{a+b}{4a+3}+\frac{b+c}{4b+3}+\frac{c+a}{4c+3}\le \frac{6}{7},$$which is wrong.

I found countered example $a=\dfrac{\sqrt{390}}{20};b=\dfrac{9}{20};c=\dfrac{27}{20}$ that $$\frac{a+b}{4a+3}+\frac{b+c}{4b+3}+\frac{c+a}{4c+3}-\frac{6}{7}\approx 0.00295.$$ Hope to see some better ideas. Thank you.

Answer 1

By C-S $$\sum_{cyc}\sqrt{\frac{a+b}{4a+3}}\leq\sqrt{\sum_{cyc}\frac{a+b}{(4a+3)(4a+8b+3c)}\sum_{cyc}(4a+8b+3c)}$$ and it's enough to prove that: $$\sum_{cyc}\frac{a+b}{(4a+3)(4a+8b+3c)}\leq\frac{6}{35(a+b+c)},$$ which is true, but my proof of the last inequality is not so nice.