Proving $5(a^2+b^2+c^2)+ab+bc+ca\ge 2\sum_{cyc}a\sqrt{5a^2+b^2+c^2+ab+ac}.$

Question

For all $a,b,c\ge 0$ prove that$$\color{black}{5(a^2+b^2+c^2)+ab+bc+ca\ge 2\sum_{cyc}a\sqrt{5a^2+b^2+c^2+ab+ac}.}$$ I've tried to use AM-GM without success.

Indeed, $$3\cdot RHS=2\sum_{cyc}3a\sqrt{5a^2+b^2+c^2+ab+ac}\le \sum_{cyc}\left(9a^2+5a^2+b^2+c^2+ab+ac\right)=16(a^2+b^2+c^2)+2(ab+bc+ca) $$ It remains to prove $$15(a^2+b^2+c^2)+3(ab+bc+ca)\ge16(a^2+b^2+c^2)+2(ab+bc+ca) $$ $$\iff a^2+b^2+c^2\le ab+bc+ca.$$ It is obviously wrong since the last inequality is equivalent to $$(a-b)^2+(b-c)^2+(c-a)^2\le 0.$$ Could you please help me prove the starting inequality? Thank you for your help.

Answer 1

Proof.

By using AM-GM as \begin{align*} 2\sqrt{5a^2+b^2+c^2+ab+ac}&=2\sqrt{\frac{5a^2+b^2+c^2+ab+ac}{\sqrt{bc}+2a}\cdot\left(\sqrt{bc}+2a\right)}\\&\le \frac{5a^2+b^2+c^2+ab+ac}{\sqrt{bc}+2a}+\sqrt{bc}+2a\\&=\frac{bc+a^2+ab+ac+b^2+c^2}{\sqrt{bc}+2a}+4a. \end{align*}Also,$$2a+\sqrt{bc}\ge 2a+\frac{2bc}{b+c}=\frac{2(ab+bc+ca)}{b+c}.$$ Subsequently, $$\frac{(ab+ac)[bc+ab+ac+a^2+b^2+c^2]}{2(ab+bc+ca)}+4a^2\ge 2a\sqrt{5a^2+b^2+c^2+ab+ac}.$$ Sum up analogs, we get desired inequality. Equality holds iff $a=b=c.$

Answer 2

The last inequality is not right. You can multiply both sides of the inequality by $2(a^2b^2+b^2c^2+c^2a^2)$ and expand it, and then you will find that the direction of the inequality sign is opposite.

Answer 3

Bacteria forever!

By C-S $$\sum_{cyc}a\sqrt{5a^2+b^2+c^2+ab+ac}\leq\sqrt{\sum_{cyc}\frac{a}{b+c}\sum_{cyc}a(b+c)(5a^2+b^2+c^2+ab+ac)}$$ and it's enough to prove that: $$\sum_{cyc}(5a^2+ab)\geq2\sqrt{\sum_{cyc}\frac{a}{b+c}\sum_{cyc}a(b+c)(5a^2+b^2+c^2+ab+ac)}$$ and the rest is smooth.

Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that: $$45u^2-27v^2\geq2\sqrt{\frac{27u^3-18uv^2+3w^3}{9uv^2-w^3}\cdot(162u^2v^2-90v^4-18uw^3)}$$ or $f(w^3)\geq0$, where $$f(w^3)=3(5u^2-3v^2)^2(9uv^2-w^3)-8(9u^3-6uv^2+w^3)(9u^2v^2-5v^4-uw^3).$$ But $$f'(w^3)=-3(5u^2-3v^2)^2-8(9u^2v^2-5v^4-uw^3)+8(9u^3-6uv^2+w^3)u=$$ $$=-(3u^4+30u^2v^2-13v^4-169w^3)\leq0,$$ which says that $f$ decreases and it's enough to prove $f(w^3)\geq0$ for a maximal value of $w^3$, which by $uvw$ happens for equality case of two variables.

Since $f(w^3)\geq0$ is homogeneous, it's enough to check $b=c=1$(for $b=c=0$ this inequality is obviously true), which gives: $$5a^2+2a+11\geq2\sqrt{\left(\frac{a}{2}+\frac{2}{a+1}\right)\left(2a(5a^2+2+2a)+2(a+1)(a^2+7+a)\right)}$$ or $$(a-1)^2(a^3+7a^2-5a+9)\geq0$$ and we are done!