# Prove $\frac{a}{\sqrt{4a+3bc}}+\frac{b}{\sqrt{4b+3ca}}+\frac{c}{\sqrt{4c+3ab}}\le \frac{\sqrt{a+b+c+2}}{2}.$

Problem. Let $$a,b,c\ge 0: ab+bc+ca=1.$$ Prove that $$\frac{a}{\sqrt{4a+3bc}}+\frac{b}{\sqrt{4b+3ca}}+\frac{c}{\sqrt{4c+3ab}}\le \frac{\sqrt{a+b+c+2}}{2}.$$ Equality holds at $$abc=0.$$

I've tried to use Cauchy-Schwarz inequality $$\sum_{cyc}\sqrt{\frac{a}{4a+3bc}}\cdot \sqrt{a}\le \sqrt{(a+b+c)\cdot \left(\frac{a}{4a+3bc}+\frac{b}{4b+3ca}+\frac{c}{4c+3ab}\right)},$$ hence it's enough to prove $$\frac{a}{4a+3bc}+\frac{b}{4b+3ca}+\frac{c}{4c+3ab}\le \frac{a+b+c+2}{4(a+b+c)}.$$The last inequality is wrong when $$a=0$$.

I'd like to ask a better approach. All idea and comment are welcome.

By C-S $$\sum_{cyc}\frac{a}{\sqrt{4a+3bc}}=\sqrt{\sum_{cyc}\frac{a^2}{4a+3bc}+2\sum_{cyc}\frac{ab}{\sqrt{(4a+3bc)(4b+3ac)}}}\leq$$ $$\leq\sqrt{\sum_{cyc}\frac{a^2}{4a+3bc}+2\sqrt{\sum_{cyc}ab\sum_{cyc}\frac{ab}{(4a+3bc)(4b+3ac)}}}$$ and it's enough to prove that: $$\sum_{cyc}\frac{a^2}{4a+3bc}+2\sqrt{\sum_{cyc}\frac{ab}{(4a+3bc)(4b+3ac)}}\leq\frac{a+b+c+2}{4}.$$ Now, let $$a+b+c=3u$$, $$ab+ac+bc=3v^2$$, where $$v=\frac{1}{\sqrt3}$$ and $$abc=w^3$$.

Thus, we need to prove that: $$\frac{3u+2}{4}-\frac{\sum\limits_{cyc}a^2(4b+3ac)(4c+3ab)}{\prod\limits_{cyc}(4a+3bc)}\geq2\sqrt{\frac{\sum\limits_{cyc}ab(4c+3ab)}{\prod\limits_{cyc}(4a+3bc)}}$$ or $$648u^2w^3+81uw^6-468uw^3-54w^6+32w^3+96\geq8\sqrt{3(4w^3+1-6uw^3)(324u^2w^3-288uw^3+27w^6-8w^3+48)}$$ or after cancelation of $$w^3$$ we need to prove that $$f(u)\geq0,$$ where $$f(u)=419904u^4w^3+104976u^3w^6-233280u^3w^3+6561u^2w^9-145800u^2w^6-$$ $$-320112u^2w^3+62208u^2-8748uw^9+86832uw^6+197568uw^3+20736u+$$ $$+2916w^9-24192w^6-8384w^3-29187.$$ But $$f'(u)=1259712u^3w^3+314928u^2w^6-699840u^2w^3+13122uw^9-291600uw^6-$$ $$-640224uw^3+124416u-8748w^9+86832w^6+197568w^3+20736.$$ But $$1259712u^3w^3-699840u^2w^3=1259712u^3w^3-699840\sqrt3u^2vw^3>47553u^3w^3,$$ $$124416u-640224uw^3=124416\cdot3\sqrt3uv^3-640224uw^3>6260uv^3,$$ $$13122uw^9-8748w^9=13122uw^9-8748\sqrt3vw^9>-2030vw^9$$ and $$314928u^2w^6-291600uw^6=314928u^2w^6-291600\cdot\sqrt3uvw^6>-190139uvw^6,$$ which says $$f'(u)>47553u^3w^3-190139uvw^6-2030uw^9+6260uv^3+86832w^6+197568w^3+20736>20736-190139uvw^6-2030uw^9=20736\cdot81v^8-190139uvw^6-2030uw^9>0.$$ Thus, $$f$$ increases and by $$uvw$$ it's enough to check $$f(u)\geq0$$ for equality case of two variables.

Let $$b=a$$ and $$c=\frac{1-a^2}{2a},$$ where $$0

Thus, we need to prove that: $$24+88a-39a^2-52a^3+138a^4-36a^5-27a^6\geq$$ $$\geq16\sqrt{6a(3a^4-4a^3-2a^2+4a+1)(3a^3-2a^2+2)}$$ or $$(1-a^2)\left(576+1152a-58a^2+1008a^3+13129a^4-6168a^5-5643a^6+5184a^7+5427a^8-1944a^9-729a^{10}\right)\geq0,$$ which is obvious for $$0

• Thank you for help. I tried to square both side but it's quite complicated. Commented Nov 3, 2023 at 1:16
• @Anonymous It's a bit of complicated, but it gives a proof. I am ready to show a full solution. Commented Nov 3, 2023 at 5:20
• @Dragon boy The equality occurs for $abc=0$. Commented Nov 4, 2023 at 10:50
• @MichaelRozenberg Oh, you're right. Commented Nov 4, 2023 at 22:34
• @MichaelRozenberg Thanks again. I've already accepted it. Commented Nov 7, 2023 at 7:31