Prove that $\frac{6(a^2 + b^2 + c^2)}{a + b + c} \geq \frac{(a + b)^2}{b + c} + \frac{(b + c)^2}{c + a} + \frac{(a + c)^2}{a + b}$

Question

Prove that if $a,b,c$ are the lengths of the edges of a given triangle, then the following inequality holds:

$\frac{6(a^2 + b^2 + c^2)}{a + b + c} \geq \frac{(a + b)^2}{b + c} + \frac{(b + c)^2}{c + a} + \frac{(a + c)^2}{a + b}$.

I have tried so far using the Ravi substitution, but the expression got even uglier. I have also tried making the substitutions: $x = a + b$, $y = b + c$, $z = a + c$, and after using $3(a^2 + b^2 + c^2) \geq (a + b + c)^2$, this resulted in showing that: $x + y + z \geq \frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x}$, where we also know (from triangle inequality) that: $z \geq \frac{x+y}{3}$ and the other 2 analogues, but at this point I got stuck.

Answer 1

Computer assisted proof.

The sides of the triangle can be cyclically rearranged, so that $a$ is the smallest side. Then the other two sides are of the shape $a+s$, $a+s+t$ with $0\le s$ and $0\le t<a$. (The last inequality, $t< a$, insures that the biggest side, $a+s+t$, is $< a+(a+s)$.

We have then to consider the two cases, where $a+s$, and $a+s+t$ are either $b,c$, or $c,b$.

We compute the difference of the quantities in the OP:

sage: var('a,b,c,s,t');
sage: f(a,b,c) = (a+b)^2 / (b+c)
sage: E(a,b,c) = 6*(a^2+b^2+c^2)/(a+b+c) - f(a,b,c) - f(b,c,a) - f(c,a,b)
sage: G(a,b,c) = E(a,b,c).numerator()

sage: G(a,a+s,a+s+t).expand()
8*a^3*s^2 + 18*a^2*s^3 + 11*a*s^4 + 2*s^5 + 8*a^3*s*t 
+ 36*a^2*s^2*t + 34*a*s^3*t + 9*s^4*t + 8*a^3*t^2 + 30*a^2*s*t^2
+ 33*a*s^2*t^2 + 10*s^3*t^2 + 6*a^2*t^3 + 10*a*s*t^3 + 3*s^2*t^3
- a*t^4 - 2*s*t^4 - t^5

sage: G(a,a+s+t,a+s).expand()
8*a^3*s^2 + 18*a^2*s^3 + 11*a*s^4 + 2*s^5 + 8*a^3*s*t 
+ 18*a^2*s^2*t + 10*a*s^3*t + s^4*t + 8*a^3*t^2 + 12*a^2*s*t^2 
- 3*a*s^2*t^2 - 6*s^3*t^2 
+ 6*a^2*t^3
- 2*a*s*t^3 - 7*s^2*t^3 - a*t^4 - 4*s*t^4 - t^5

(Sage code was inserted. Slightly rearranged to fit in the window.)

So we have to dominate the minus terms by using the plus terms.

In the first case things are simple, we have for instance $8a^3t^2 -at^4-t^5>0$, and $8a^3st-2st^4>0$.

In the second case we have more to type, there is no intelligence involved more than matching the degrees of $s$ while covering the minus terms with the plus terms: $$ \begin{aligned} 8 \, a^{3} s^{2} + 18 \, a^{2} s^{2} t - 3 \, a s^{2} t^{2} - 7 \, s^{2} t^{3} &\ge 0\ ,\\ 18 \, a^{2} s^{3} - 6 \, s^{3} t^{2} &\ge 0\ ,\\ 8 \, a^{3} s t - 2 \, a s t^{3} - 4 \, s t^{4} &\ge 0\ ,\\ 8 \, a^{3} t^{2} + 6 \, a^{2} t^{3} - a t^{4} - t^{5} &>0\ . \end{aligned} $$ $\square$

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Personal note: The given proof misses by intention any beauty, since the author of the problem also tried desperately to offer an inequality without symmetry, without a visual possibility of a Chebyshev rearrangement, and with the $\ge$ sign on the side that does not allow an application of a Cauchy inequality, so that things may become symmetric after some steps. I hope (s)he may enjoy this proof, at least as i enjoyed typing it.

Answer 2

Let $c=\max\{a,b,c\}$, $a=x+u$, $b=x+v$ and $c=x+u+v$, where $x>0$, $u\geq0$ and $v\geq0.$

Thus, we need to prove that $$8(u^2-uv+v^2)x^3+6(3u^3-2u^2v+uv^2+3v^3)x^2+$$ $$+(11u^4+2u^3v-15u^2v^2+26uv^3+11v^4)x+2(u^5+u^4v-3u^3v^2+u^2v^2+5uv^4+v^5)\geq0,$$ which is obvious.

Also, the $uvw$'s technique helps, but it's very complicated here.