For $a,b,c > 0$ prove:
$$(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) +\frac{486(ab+bc+ca)^3}{(a+b+c)^6} \geqq 27$$
My work:
I can easy found SOS for it:
$$\text{LHS-RHS}=\sum {\frac { \left( a-b \right) ^{2}\cdot M}{{a}^{2}{b}^{2} \left( a+b+c \right)
^{6}}} \geqq 0$$
Where $M=\left( a+b \right) ^{2}{c}^{6}+6\, \left( a+b \right) ^{3}{c}^{5}$
$+
\left( 15\,{a}^{4}+60\,{a}^{3}b+81\,{a}^{2}{b}^{2}+60\,a{b}^{3}+15\,{
b}^{4} \right) {c}^{4}$
$+ \left( a+b \right) \left( 20\,{a}^{4}+80\,{a}
^{3}b+57\,{a}^{2}{b}^{2}+80\,a{b}^{3}+20\,{b}^{4} \right) {c}^{3}$
$+
\left( 15\,{a}^{6}+90\,{a}^{5}b+36\,{a}^{4}{b}^{2}-105\,{a}^{3}{b}^{3
}+36\,{a}^{2}{b}^{4}+90\,a{b}^{5}+15\,{b}^{6} \right) {c}^{2}$
$+3\,
\left( a+b \right) \left( 2\,{a}^{6}+12\,{a}^{5}b+9\,{a}^{4}{b}^{2}-
74\,{a}^{3}{b}^{3}+9\,{a}^{2}{b}^{4}+12\,a{b}^{5}+2\,{b}^{6} \right) c
$
$+ \left( {a}^{6}+3\,{a}^{5}b+3\,{a}^{4}{b}^{2}-25\,{a}^{3}{b}^{3}+3\,{
a}^{2}{b}^{4}+3\,a{b}^{5}+{b}^{6} \right) \left( {a}^{2}+5\,ab+{b}^{2
} \right) \geqq 0$
But how to prove $M\geqq 0$$?$ Then I struck here.
Plan text for M:
M :=a^8 + 8*a^7*b + 6*a^7*c + 19*a^6*b^2 + 42*a^6*b*c + 15*a^6*c^2 - 7*a^5*b^3 + 63*a^5*b^2*c + 90*a^5*b*c^2 + 20*a^5*c^3 - 119*a^4*b^4 - 195*a^4*b^3*c + 36*a^4*b^2*c^2 + 100*a^4*b*c^3 + 15*a^4*c^4 - 7*a^3*b^5 - 195*a^3*b^4*c - 105*a^3*b^3*c^2 + 137*a^3*b^2*c^3 + 60*a^3*b*c^4 + 6*a^3*c^5 + 19*a^2*b^6 + 63*a^2*b^5*c + 36*a^2*b^4*c^2 + 137*a^2*b^3*c^3 + 81*a^2*b^2*c^4 + 18*a^2*b*c^5 + a^2*c^6 + 8*a*b^7 + 42*a*b^6*c + 90*a*b^5*c^2 + 100*a*b^4*c^3 + 60*a*b^3*c^4 + 18*a*b^2*c^5 + 2*a*b*c^6 + b^8 + 6*b^7*c + 15*b^6*c^2 + 20*b^5*c^3 + 15*b^4*c^4 + 6*b^3*c^5 + b^2*c^6
PS: I found this inequality when I try to use AM-GM to prove this inequality:
$$(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+18\cdot \frac{ab+bc+ca}{a^2+b^2+c^2}\geqq 27$$
$\lceil $See also here: https://artofproblemsolving.com/community/c6h2086137p15058647 $\rfloor$
The following stronger inequality is also true!
$$(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) +\frac{k(ab+bc+ca)^3}{(a+b+c)^6} \geqslant 9+\frac{1}{27}k$$
where $k\approx 618.6094263$ is a root of
${k}^{6}-{\frac {26032158}{50653}}\,{k}^{5}+{\frac {126036095580}{1369}
}\,{k}^{4}-{\frac {3283611347814696}{50653}}\,{k}^{3}$
$+{\frac {
274967018226970704}{50653}}\,{k}^{2}-{\frac {18251898690181651200}{
50653}}\,k+{\frac {491942544951481344}{50653}}=0
$
My software say this is the maximum value of k, but I have no proof for it. Who can?
Best Answer
It's impossible because for $c\rightarrow0^+$ we get that $M$ can be negative: try $a=b=1$.
But $uvw$ kills your inequality immediately!
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that $$\frac{(9u^2-6v^2)(9v^4-6uw^3)}{w^6}+\frac{486\cdot27v^6}{729u^6}\geq27$$ or $$\frac{(3u^2-2v^2)(3v^4-2uw^3)}{w^6}+\frac{2v^6}{u^6}\geq3$$ or $f(w^3)\leq0$ where $$f(w^3)=\left(3-\frac{2v^6}{u^6}\right)w^6-(3u^2-2v^2)(3v^4-2uw^3).$$ But $f$ is a convex function and the convex function gets a maximal value
for an extreme value of $w^3$, which happens for equality case of two variables
(the case $w^3\rightarrow0^+$ is trivial).
Since our inequality is homogeneous, we can assume $b=c=1$ and we need to prove that $$(a^2+2)\left(\frac{1}{a^2}+2\right)+\frac{486(2a+1)^3}{(a+2)^6}\geq27$$ or $$(a^2+2)\left(\frac{1}{a^2}+2\right)-9\geq18-\frac{486(2a+1)^3}{(a+2)^6}$$ or $$\frac{2(a-1)^2(a+1)^2}{a^2}\geq\frac{18(a-1)^2(a^4+14a^3+87a^2+104a+37)}{(a+2)^6}.$$ But by AM-GM $$(a+1)^2\geq4a$$ and $$a^4+14a^3+87a^2+104a+37\leq(a+2)^4.$$ Id est, it's enough to prove that $$\frac{4}{a}\geq\frac{9}{(a+2)^2},$$ which is true by AM-GM again: $$\frac{9}{(a+2)^2}\leq\frac{9}{(2\sqrt{2a})^2}=\frac{9}{8a}<\frac{4}{a}$$ and we are done!
The following stronger inequality is also true.