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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?

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5 Answers 5

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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.

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) +\frac{594(ab+bc+ca)^3}{(a+b+c)^6} \geq 31$$

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    $\begingroup$ Yeah! Show it's please! I hope for $uvw$ solution! $\endgroup$
    – NKellira
    May 6, 2020 at 3:48
  • $\begingroup$ Thank you! Could you have a nice SOS for it? I can't! $\endgroup$
    – NKellira
    May 6, 2020 at 4:34
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    $\begingroup$ @tthnew I'll try. I think it's possible. But I am sure that it's impossible for my inequality, which I posted. $\endgroup$ May 6, 2020 at 4:37
  • $\begingroup$ Wow really? Degree of these inequality is high, so it is not easy to find nice SOS with me. $\endgroup$
    – NKellira
    May 6, 2020 at 4:39
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    $\begingroup$ @tthnew I proved your inequality by SOS. See please my solution. Also, I proved my inequality by SOS and now I think we can get by SOS something more stronger. $\endgroup$ May 6, 2020 at 6:11
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A proof by SOS.

We need to prove that: $$\sum_{cyc}\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}-2\right)\geq18-\frac{486(ab+ac+bc)^3}{(a+b+c)^6}$$ or $$\sum_{cyc}\frac{(a^2-b^2)^2}{a^2b^2}\geq\frac{18((a+b+c)^6-27(ab+ac+bc)^3)}{(a+b+c)^6}$$ or $$\sum_{cyc}\frac{(a^2-b^2)^2}{a^2b^2}\geq$$ $$\geq\tfrac{18((a+b+c)^2-3(ab+ac+bc))((a+b+c)^4+3(a+b+c)^2(ab+ac+bc)+9(ab+ac+bc)^2)}{(a+b+c)^6}$$ or $$\sum_{cyc}(a-b)^2\left(\frac{(a+b)^2}{a^2b^2}-\tfrac{9((a+b+c)^4+3(a+b+c)^2(ab+ac+bc)+9(ab+ac+bc)^2)}{(a+b+c)^6}\right)\geq0$$ and since $$(a+b+c)^2\geq3(ab+ac+bc)$$ and by AM-GM $$(a+b)^2\geq4ab,$$ it's enough to prove that: $$\sum_{cyc}(a-b)^2\left(\frac{4}{ab}-\tfrac{27}{(a+b+c)^2}\right)\geq0$$ or $$\sum_{cyc}(a-b)^2c\left((a+b+c)^2-\frac{27}{4}ab\right)\geq0$$ for which it's enough to prove that: $$\sum_{cyc}(a-b)^2c\left((a+b+c)^2-7ab\right)\geq0.$$ Now, let $a\geq b\geq c$.

Thus, by AM-GM $$(a+b+c)^2-7ac\geq\left(2\sqrt{a(b+c)}\right)^2-7ac\geq\left(2\sqrt{2ac}\right)^2-7ac=ac>0,$$ which by AM-GM again gives $$\sum_{cyc}(a-b)^2c\left((a+b+c)^2-7ab\right)\geq$$ $$\geq(a-b)^2c\left((a+b+c)^2-7ab\right)+ (a-c)^2b((a+b+c)^2-7ac)\geq$$ $$\geq(a-b)^2c\left((a+b+c)^2-7ab\right)+ (a-b)^2b((a+b+c)^2-7ac)=$$ $$=(a-b)^2((b+c)(a+b+c)^2-14abc)\geq(a-b)^2\left(2\sqrt{bc}(a+2\sqrt{bc})^2-14abc\right)\geq$$ $$\geq(a-b)^2\left(2\sqrt{bc}\left(2\sqrt{a\cdot2\sqrt{bc}}\right)^2-14abc\right)=2abc(a-b)^2\geq0$$ and we are done!

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    $\begingroup$ Thank you! I find it: $$\sum_{cyc}(a-b)^2c \left((a+b+c)^2-7ab\right) =\sum\limits_{cyc} c \left( a-b \right) ^{2} \left( \frac{1}{8}\, \left( -6\,c+a+b \right) ^{2}+{ \frac {7\, \left( a-b \right) ^{2}}{8}} \right)\geqq 0 $$ $\endgroup$
    – NKellira
    May 6, 2020 at 6:10
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    $\begingroup$ @tthnew I think there is a mistake in your computations. Try $c\rightarrow0$ and $a=b=1$. In this case $(a+b+c)^2-7ab<0$. $\endgroup$ May 6, 2020 at 6:13
  • $\begingroup$ I was used computer to check it, no error found. $\endgroup$
    – NKellira
    May 6, 2020 at 6:15
  • $\begingroup$ Of course $$(a+b-c)^2 -7ab \neq \frac{1}{8}(a+b-6c)^2 +\frac{7}{8}{(a-b)^2}$$ but $\sum_{cyc}(a-b)^2c \left((a+b+c)^2-7ab\right) =\sum\limits_{cyc} c \left( a-b \right) ^{2} \left( \frac{1}{8}\, \left( -6\,c+a+b \right) ^{2}+{ \frac {7\, \left( a-b \right) ^{2}}{8}} \right)$ $\endgroup$
    – NKellira
    May 6, 2020 at 6:16
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    $\begingroup$ @tthnew OK. I agree we can find it by hand. I checked your identity. It's really amazing and true! But I think, it's better to end the proof by my way. What do you think? $\endgroup$ May 6, 2020 at 6:28
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The inequality $(a^2 + b^2 + c^2)(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}) + 18 \cdot 27 \frac{(a b + b c + a c)^3}{(a+b+c)^6}- 27 \ge 0$ can be proved by the standard substitution $a= u$, $b=u+v$, $c= u+v+w$. In fact, by the same method one can show the stronger inequality $$ (a^2 + b^2 + c^2)(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}) + 21.6 \cdot 27 \frac{(a b + b c + a c)^3}{(a+b+c)^6}- 30.6 \ge 0$$ This is as far as we can go with this method.

An equivalent statement is that the function $ f_t(a,b,c)=(a^2 + b^2 + c^2)(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}) + t \frac{(a b + b c + a c)^3}{(a+b+c)^6}$ has a minimum ( say on unit simplex $a,b,c\ge 0$, $a+b+c=1$) at $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$. Increasing $t$ we get a stronger inequality. How far can we increase $t$? It is clear that we cannot increase indefinitely since the second term has a maximum at $(\frac{1}{3},\frac{1}{3},\frac{1}{3})$. Numerics suggest that the minimum is achieved in any case at points with at least two equal coordinates. Therefore, to check how far we can go with $t$, we consider the function $g_t(a) = f_t(a,1,1)$ and we ask what is the largest $t$ for which the minimum is achieved at $a=1$. For this, we look at the zeroes of $\frac{d g_t}{d a}$, there are two of them, one at $a=1$, and one at a value of $a$ that depends implicitly on $t$. Now postulate that the value of $g_t$ at these two critical points are equal. Solving the system of two equations in $t$, $a$, get a value of $t$ approximately $t = 618.609...$. It is worth comparing the Minimize and NMinimize commands with Mathematica for $t$ close to this critical value.

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    $\begingroup$ Yes, BW kills this inequality. But can you prove it with this way by hand? Also, see please my post, where I proved this inequality by $uvw$. My inequality there is stronger than your. $\endgroup$ May 6, 2020 at 5:24
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    $\begingroup$ @Michael Rozenberg: It is interesting that some of the stronger inequalities cannot be proved by the substitution. Only checked them numerically, so, no proof, $\endgroup$
    – orangeskid
    May 6, 2020 at 5:59
  • $\begingroup$ I add something about the max value, see now. $\endgroup$
    – NKellira
    Aug 13, 2020 at 9:36
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The pqr method:

WLOG, assume that $a+b+c = 3$.

Let $p = a+b+c = 3$, $q = ab+bc+ca$ and $r = abc$.

We have the following known facts: $q^2 \ge 3pr$, $p^2 \ge 3q$ and $q^3 + 9r^2 \ge 4pqr$.
Remark: The last one is nothing but $ab(ab-bc)(ab-ca) + bc(bc-ca)(bc-ab) + ca(ca-ab)(ca-bc) \ge 0$ (Schur's inequality for $ab, bc, ca$).

From $q^3 + 9r^2 \ge 4pqr$, we have $(\frac{1}{r} - \frac{2p}{q^2})^2\ge \frac{4p^2-9q}{q^4}$ and $\frac{1}{r} - \frac{2p}{q^2} \ge \frac{\sqrt{4p^2-9q}}{q^2}$ (note: $\frac{1}{r} - \frac{2p}{q^2} = \frac{q^2-2pr}{rq^2} > 0$). Thus, we have \begin{align} \frac{1}{a^2}+ \frac{1}{b^2} + \frac{1}{c^2} &= \frac{q^2 - 2pr}{r^2}\\ &= q^2\left(\frac{1}{r} - \frac{p}{q^2}\right)^2 - \frac{p^2}{q^2}\\ &\ge q^2\left(\frac{p + \sqrt{4p^2-9q}}{q^2}\right)^2 - \frac{p^2}{q^2}. \end{align} It suffices to prove that $$(p^2-2q)\left(q^2\left(\frac{p + \sqrt{4p^2-9q}}{q^2}\right)^2 - \frac{p^2}{q^2}\right) + \frac{486q^3}{p^6} - 27 \ge 0$$ or $$(9 - 2q)\left(\frac{(3 + 3\sqrt{4-q})^2}{q^2} - \frac{9}{q^2}\right) + \frac{2}{3}q^3 - 27 \ge 0.$$ From $p^2 \ge 3q$, we have $q \le 3$. With the substitution $q = 4 - x^2$ for $1 \le x < 2$, it suffices to prove that, for $1\le x < 2$, $$\frac{1}{3(x-2)^2(x+2)}(-2x^7+34x^5+4x^4-218x^3-56x^2+591x+376)(x-1)^2\ge 0$$ which is true. We are done.

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  • $\begingroup$ You idea to get $(\frac{1}{r} - \frac{2p}{q^2})^2\ge \frac{4p^2-9q}{q^4}$ is nice! $\endgroup$
    – NKellira
    Aug 14, 2020 at 0:57
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    $\begingroup$ @tthnew For example, $f = (a+b)(b+c)(c+a)$, I want to find its pqr expression. Consider $f_1 = f - x$, $f_2 = a + b + c - p$, $f_3 = ab + bc + ca - q$ and $f_4 = abc - r$. We calculate the resultants $\mathrm{Res}(f_1, f_2, c) = g_1^m$, $\mathrm{Res}(f_2, f_3, c) = g_2^n$ and $\mathrm{Res}(f_3, f_4, c) = g_3^k$ where $m, n, k$ are positive integers, then calculate $\mathrm{Res}(g_1, g_2, b) = h_1^M$ and $\mathrm{Res}(g_2, g_3, b) = h_2^N$ where $M, N$ are positive integers. $\endgroup$
    – River Li
    Aug 22, 2020 at 10:12
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    $\begingroup$ @tthnew Finally, calculate $\mathrm{Res}(h_1, h_2, a) = F(x, p, q, r)^L$ ($L$ is a positive integer) and we will get the pqr expression by solving $F(x, p, q, r)=0$ to get $x$. (If $f$ is symmetric, $F(x, p, q, r)$ is affine in $x$). $\endgroup$
    – River Li
    Aug 22, 2020 at 10:12
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    $\begingroup$ @tthnew Yes, it works well. You are good at writing Maple code. $\endgroup$
    – River Li
    Aug 22, 2020 at 12:50
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    $\begingroup$ @tthnew For asymmetric and acyclic polynomials, for example, $x = a + 2b + c$, we have $-x^3+4px^2+(-5p^2-q)x+2p^3+qp+r = 0$. $\endgroup$
    – River Li
    Aug 29, 2020 at 1:39
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About your last problem.

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

Thus, we need to find a maximal value of $k$ for which the inequality $f(w^3)\geq0,$ where $$f(w^3)=(3u^2-2v^2)(3v^4-2uw^3)-\left(1+k\left(\frac{1}{243}-\frac{v^6}{243u^6}\right)\right)w^6$$ is true for any positives $a$, $b$ and $c$.

But since $$f'(w^3)=-2u(3u^2-2v^2)-2\left(1+k\left(\frac{1}{243}-\frac{v^6}{243u^6}\right)\right)w^3<0,$$ we see that $f$ decreases, which says that $f$ gets a minimal value for a maximal value of $w^3$,

which happens for equality case of two variables.

Since, our inequality is homogeneous and symmetric, it's enough to assume $b=c=1$, which gives $$(a^2+2)\left(\frac{1}{a^2}+2\right)-9\geq k\left(\frac{1}{27}-\frac{(a+1)^3}{(a+2)^6}\right)$$ or $$\frac{54(a+1)^2(a+2)^6}{a^2(a^4+14a^3+87a^2+104a+37)}\geq k,$$ which gives $$k_{max}=\min_{a>0}\frac{54(a+1)^2(a+2)^6}{a^2(a^4+14a^3+87a^2+104a+37)}=618.6...$$ Indeed, $$\left(\frac{54(a+1)^2(a+2)^6}{a^2(a^4+14a^3+87a^2+104a+37)}\right)'=$$ $$=\frac{108(a+1)(a+2)^5(a^6+17a^5+133a^4+103a^3-185a^2-238a-74)}{a^3(a^4+14a^3+87a^2+104a+37)^2}.$$ We see that by the Descartes's rule the polynomial $$a^6+17a^5+133a^4+103a^3-185a^2-238a-74$$ has an unique positive root and easy to see that this root ($\approx 1.317...$) gives a minimal value.

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  • $\begingroup$ Thanks, that problem is hard with me, I create it by Maple but can not solve it. $\endgroup$
    – NKellira
    Aug 13, 2020 at 23:00

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