Solving $(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge(3+2\sqrt{3})^2$

Question

Given $a,b,c$ are real numbers such that: $abc\neq0$ and $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3.$$ Prove that: $$(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge(3+2\sqrt{3})^2$$

The problem is quite hard for me because I don't know where equality occur.

Baically, the symmetric inequality become equality when $a=b=c$. Unfortunately,this statement makes no sense since

$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=9$$

I've tried to let $$A=a+b+c; B=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}; C=ab+bc+ca;D=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}$$ the inequality turns out $$(A^2-2C)(B^2-2D)\ge(3+2\sqrt{3})^2$$

I replace $AB=3$ but I don't know what is the rest.

Hope you can help me. Thank you.

One minor observation is that, if $(a,b,c)$ satisfies the constraint equation, then so too does $(\lambda a,\lambda b,\lambda c)$ where $\lambda\neq 0$. This allows one to assume, for instance, that $a+b+c=1$ without loss of generality. (Or $a^2+b^2+c^2=1$ for that matter---it is not obvious to me which if either of these is more useful.) — Semiclassical, Dec 19 '23 at 05:00
By setting $ a = 1$, you can determine the 6 equality cases. — Calvin Lin, Dec 19 '23 at 05:26
@CalvinLin It seems that $$(a,b,c)\sim \left(t;kt;-t\sqrt{k}\right)$$ where $t\neq 0; k\in{1+\sqrt{3}+\sqrt{3+2\sqrt{3}};1+\sqrt{3}-\sqrt{3+2\sqrt{3}}}.$ — TATA box, Dec 19 '23 at 08:02

Macavity · Answer 1 · 2023-12-19T16:33:42.123

Just another way: Let $p = \sum_{cyc} \frac{a}b, q = \sum_{cyc} \frac{b}a$. The given condition then is $p+q=0$.

We have $pq = 3 + r + s$, where $r = \sum_{cyc} \frac{ab}{c^2}, s = \sum_{cyc} \frac{c^2}{ab}$.
Further, non-zero $p+q=0\implies pq<0 \implies r+s < -3$. We also have $$rs = 3 + \sum_{sym}\frac{a^3}{b^3}$$ Now $p+q =0 \implies p^3+q^3=0 \implies$ $$p^3+q^3=\sum_{sym}\frac{a^3}{b^3}+12+6(r+s)=0 \implies rs=-9-6(r+s)$$

Thus $r, s$ are real solutions to the quadratic $x^2-\lambda x-(9+6\lambda)$ where $\lambda = r+s < -3$. From the discriminant of the quadratic, we must have $\lambda \leqslant -6(2+\sqrt3)$, therefore $pq = 3+r+s \leqslant -9-6\sqrt3$. Hence we finally get

$$\sum_{cyc} a^2\cdot \sum_{cyc}\frac1{a^2} = 3+\sum_{cyc}\frac{a^2}{b^2}+ \sum_{cyc}\frac{b^2}{a^2}=3+(p^2-2q)+(q^2-2p)\\=3+p^2+q^2 =3-2pq \geqslant 21+12\sqrt3=(3+2\sqrt3)^2$$

P.S. While not asked for in the problem, with some work, it can be verified that this inequality becomes an equality for $(a, b, c)\sim (2, -1-\sqrt3 -\sqrt2\sqrt[4]3,-1-\sqrt3 +\sqrt2\sqrt[4]3) $ or any non-zero scaled permutation thereof, hence this is indeed the minimum.

A sad math boy · Answer 2 · 2023-12-19T11:49:32.913

4

Well here's my solution.

We can assume $abc=1$ without loss of generality.

Then we notice that $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=(a+b+c)^2-2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2-2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})=(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2-2(a+b+c)$$

So if we multiply them together and expand, we get$$(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})=((a+b+c)^2-2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}))((\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2-2(a+b+c))\\=4(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+((a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}))^2-2((a+b+c)^3+(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^3)$$

Then substitute in $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{a+b+c}$, we get that the above is equal to$$21-2((a+b+c)^3+(\frac{3}{a+b+c})^3)$$

Let's make variable $x=(a+b+c)^3$,then the above is equal to $$21-2(x+\frac{27}{x})$$

Because $a^2+b^2+c^2=(a+b+c)^2-\frac{6}{a+b+c}\geq 0$,$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{9}{(a+b+c)^2}-2(a+b+c)>0$,so solving the inequalities we get $a+b+c<0$,thus $x<0$.

Then,the above is equal to $$21+2((-x)+\frac{27}{(-x)})=21+4(\frac{(-x)+\frac{27}{(-x)}}{2})\geq21+4\sqrt{27}=21+12\sqrt{3}=(3+2\sqrt3)^2$$by AM-GM.

edited Dec 19 '23 at 11:49

answered Dec 19 '23 at 07:16

A sad math boy

131

$abc=1$, how is that obvious? – Dominique Dec 19 '23 at 07:18
@Dominique Actually,see the comment by Semiclassical in the main problem – A sad math boy Dec 19 '23 at 07:20
3

Why must $ x < 0 $? I agree that not all of $a, b, c$ are positive. You'd need to hunt it down a bit more, but otherwise this looks promising. – Calvin Lin Dec 19 '23 at 07:22
@CalvinLin Yeah.I'm now wondering why x is less than zero now too. – A sad math boy Dec 19 '23 at 07:24
You should divide cases. – TATA box Dec 19 '23 at 10:01
Oh.I know why x > 0. – A sad math boy Dec 19 '23 at 10:26
@TATAbox I added it in the solution. – A sad math boy Dec 19 '23 at 10:33
$a+b+c<0$ because we always have $ab+bc+ca<0.$ (in case you assume $abc=1$) – TATA box Dec 19 '23 at 10:58
@TATAbox Why do we have that? – A sad math boy Dec 19 '23 at 11:19
@Asadmathboy Now I see you prove it. The rest is just function of $x.$ Why don't you full it? – TATA box Dec 19 '23 at 11:39
Let us continue this discussion in chat. – A sad math boy Dec 19 '23 at 11:43
Good approach +1, though I would have written it differently - typing in for reference: Using homogeneity and $abc\neq0$, WLOG we assume $abc=1$. Let $a, b, c$ be roots of $x^3-px^2+qx-1=0$. We are given $pq=3$. Further, two roots must be negative (else CS inequality would apply) and hence $p, q < 0$ from Descartes' rule of signs.
The monic cubic with roots $a^2, b^2, c^2$ is $x^3-(p^2-2q)x^2+(q^2-2p)-1$. Hence we need the minimum of $(p^2-2q)(q^2-2p)=21-2(p^3+q^3)$.

By AM-GM, $-p^3-q^3 \geqslant 2\sqrt{(pq)^3}=6\sqrt3$, so $21-2(p^3+q^3)\geqslant 21+12\sqrt3=(3+2\sqrt3)^2$
– Macavity Dec 19 '23 at 15:43

score 1 · Answer 3 · answered Dec 20 '23 at 00:45

A proof using the pqr substitution.

Remark: Sometimes, pqr substitution looks neat.

Let $p = a + b + c, q = ab + bc + ca, r = abc \ne 0$.

The condition is written as $p \cdot \frac{q}{r} = 3$ which results in $r = \frac{pq}{3}$.

We need to prove that $$(p^2 - 2q)\cdot \frac{q^2 - 2pr}{r^2} \ge (3 + 2\sqrt{3})^2$$ or (using $r = \frac{pq}{3}$) $$21 - \frac{6p^2}{q} - \frac{18q}{p^2} \ge (3 + 2\sqrt{3})^2. \tag{1}$$

We claim that $q < 0$. Indeed, clearly $a + b \ne 0$, otherwise $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) = 1 \ne 3$; Plugging $c = \frac{q - ab}{a + b}$ into $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3$, we have $$q^2 + (a - b)^2 q + 3a^2b^2= 0$$ which results in $q < 0$.

From (1) and $q < 0$, letting $x := -\frac{p^2}{q} > 0$, it suffices to prove that $$21 + 6x + \frac{18}{x} \ge (3 + 2\sqrt{3})^2$$ or $$\frac{6(x - \sqrt{3})^2}{x}\ge 0$$ which is clearly true.

We are done.

score 0 · Answer 4 · answered Dec 22 '23 at 01:29

Solution.

Let $$X=\frac{a}{b}+\frac{b}{c}+\frac{c}{a};Y=\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \implies X+Y=0$$ The inequality becomes $$X^2-2Y+Y^2-2X\ge 18+12\sqrt{3} \iff X^2\ge 9+6\sqrt{3}.$$ Or$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge 9+6\sqrt{3}.$$ WLOG, assuming that $abc > 0.$ Let $p=a + b + c, q = ab + bc + ca, r = abc>0.$ Then $pq = 3r.$

We have$$\left(\frac{p^2q^2 - 2q^3}{2r^2} + \frac{3pq - p^3}{r} - \frac{9}{2}\right) - \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2$$$$ = \frac{(a-b)(b-c)(c-a)}{2a^2b^2c^2}\left(pq-3r\right) = 0.$$ Thus, it suffices to prove that $$\frac{p^2q^2 - 2q^3}{2r^2} + \frac{3pq - p^3}{r} - \frac{9}{2} \ge 9 + 6\sqrt{3}$$ or (using $pq = 3r$) $$- \frac{3(p^2 + q\sqrt3)^2}{p^2q} \ge 0.$$ It is easy to prove that $q < 0.$ We are done!

Solving $(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge(3+2\sqrt{3})^2$

4 Answers4