crazyfehmy wrote: For all positive real numbers $a, b, c$ satisfying $ab+bc+ca \leq 1,$ prove that \[ a+b+c+\sqrt{3} \geq 8abc (\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}) \] We have \[ RHS\le \frac{16abc(a+b+c)}{(a+b)(b+c)(c+a)}\le \frac{18abc(a+b+c)}{(a+b+c)(ab+bc+ca)}\le 2\sqrt{3(ab+bc+ca)}\le LHS\] It is true because $\sqrt{3(ab+bc+ca)}\le a+b+c$ and $ab+bc+ca\le 1$ Complete proof

Since $ab+bc+ca \leqslant 1$, then $a^2 +ab +bc +ca \leqslant a^2 +1$ which rewrites as $\frac{1}{a^2+1}\leqslant \frac{1}{(a+b)(a+c)}$. Therefore \[ 8abc \left(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\right) \leqslant 8abc\left(\frac{2(a+b+c)}{(a+b)(b+c)(c+a)}\right) \] Observe that $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$, and since by AM-GM $(a+b+c)(ab+bc+ca) \geqslant 9abc$, then the famous $9(a+b)(b+c)(c+a)\geqslant 8(a+b+c)(ab+bc+ca)$, i.e. \[ \frac{18abc}{ab+bc+ca} \geqslant \frac{16abc(a+b+c)}{(a+b)(b+c)(c+a)} \] In order to prove the desired inequality, it is enough to show that $ a+b+c +\sqrt{3} \geqslant \frac{18abc}{ab+bc+ca}$ \[ \iff (a+b+c)(ab+bc+ca) +\sqrt{3}(ab+bc+ca) \geqslant 18abc\] By AM-GM $(a+b+c)(ab+bc+ca)\geqslant 9abc$, so it is enough to show that $ \sqrt{3}(ab+bc+ca) \geqslant 9abc$. By AM-GM $ab+bc+ca \geqslant 3\sqrt[3]{a^2 b^2 c^2}$, hence it's enough to show that $\sqrt{3} \geqslant 3\sqrt[3]{abc}$. Easy enough as by AM-GM $1\geqslant ab+bc+ca \geqslant 3\sqrt[3]{a^2 b^2 c^2} \implies \frac{1}{3\sqrt{3}} \geqslant abc$. Done

$a^2+1\geq a^2+ab+bc+ca=a(a+b)+c(b+a)=(a+b)(a+c)$ etc., so ${RHS=abc\sum_{cyc} \frac{1}{a^2+1}\leq \frac{16abc(a+b+c)}{(a+b)(b+c)(c+a)}\leq \frac{16abc(a+b+c)}{8abc}}=2(a+b+c)$ $\leq a+b+c+\sqrt{3}$, so it remains to prove $a+b+c\leq \sqrt{3}$, which is true because $3(ab+bc+ca)\leq 3\leq (a+b+c)^{2}$

ivanbart-15 wrote: $a+b+c\leq \sqrt{3}$, which is true because $3(ab+bc+ca)\leq 3\leq (a+b+c)^{2}$ Are you sure? Try $a=10, b=1/100, c=1/100.$

My solution First notice that $ab+bc +ca \leq 1 $ implies $a^2+1\geq (a+b)(a+c)$ Thus our inequality is equivalent to $a+b+c+ \sqrt{3} \geq \frac{16abc(a+b+c)}{(a+b)(b+c)(c+a)}$ Or $\sum a^3(b+c) + 2 \sum a^{2}b^2 +\sqrt{3}\prod(a+b) \geq 12\sum a^{2}bc$ The inequality is obvious if $a+b+c\leq \sqrt{3}$ Assume ,then, that $a+b+c\geq \sqrt{3}$ Set $x=a+b+c$ $y=ab+bc+ca$ $z=abc$ (notice that $\sum a^{3}(b+c)= (a+b+c)(\sum a^3) -(\sum a^4)$) The inequality is now reduced to $x^{2}y +xy\sqrt{3} -z\sqrt{3} \geq 17zx$ Let $f(x)=x^{2}y +xy\sqrt{3} -z\sqrt{3} - 17zx$ We have $f'(x)=2xy+y\sqrt{3}-17z>0$ Since $xy=(a+b+c)(ab+bc+ca)\geq 9z$ Therefore we only need to prove $f(x_{min})\geq 0$ which is $f(\sqrt{3})\geq 0 \Leftrightarrow 6y\geq 18\sqrt{3}z $ ,but this is certainly true since we have $y\geq y^2\geq 3xz\geq 3\sqrt{3}z$ The conclusion follows.

Seems similar to above. \[ a+b+c+\sqrt{3} \overset{?}{\geq} 8abc \left(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\right) \]\[RHS=\sum{\frac{8abc}{a^2+1}}=\sum{\frac{8abc}{a^2+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}}\leq \sum{\frac{8abc}{4\sqrt{a}.\sqrt[4]{\frac{1}{27}}}}=2\sqrt[4]{27}\sum{bc\sqrt{a}}\leq2\sum{\sqrt{bc}}\]Since $3\sqrt[3]{(abc)^2}\leq \sum{ab}\leq 1$ which yields $\frac{1}{\sqrt[4]{27}}=(\frac{1}{3})^{\frac{3}{4}}\geq \sqrt{abc}$. \[2\sum{\sqrt{bc}}=\sum{\sqrt{bc}}+\sum{\sqrt{bc}}\leq \sum{a}+\sqrt{3}=LHS\]Because $(\sum{\sqrt{bc}})^2\overset{\text{CS}}{\leq }3\sum{bc}\leq 3$ gives $\sum{\sqrt{bc}}\leq \sqrt{3}$ and $\sum{\sqrt{bc}}\leq \sum{a}$ is true by rearrangement at $(\sqrt{a},\sqrt{b},\sqrt{c})$ as desired.$\blacksquare$

For all positive real numbers $a, b, c$ satisfying $ab+bc+ca \leq 1,$ prove that \[ a+b+c+\frac{\sqrt{3}}2 \geq 6abc \left(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\right) \]