Inequality $(a+b+c)(ab+bc+ca)(a^3+b^3+c^3)\le (a^2+b^2+c^2)^3$

Question

For positive real numbers $a,b,c$ prove that $$(a+b+c)(ab+bc+ca)(a^3+b^3+c^3)\le (a^2+b^2+c^2)^3$$ My try : Rewrite this as $$\frac{a^2+b^2+c^2}{ab+bc+ca}-1\ge \frac{(a+b+c)(a^3+b^3+c^3)}{a^2+b^2+c^2)^2}-1$$ Then I am stumped, I was trying to use SOS but I think an easier proof is possible. My teacher give this to me. He combined to well known inequalities $a^2+b^2+c^2\ge ab+bc+ca$ and $(a+b+c)(a^3+b^3+c^3)\ge (a^2+b^2+c^2)^2$ and changed the sign. This is easy by uvw method but my teacher wouldn't accept it. Can some one help me? Thanks in advance.

Answer 1

Let $p=a+b+c, q=ab+bc+ca, r = abc$. Also note that $p^2\ge 3q$ and $q^2 \ge 3pr$ in general. WLOG we can set $r = 1$, so $q^2 \ge 3p$ in this case. The given inequality is then

$$p \cdot q \cdot \left(p(p^3-3q)+3 \right)\le (p^2-2q)^3$$ $$\iff p^6-(7p^4+3p)q +15 p^2 q^2-8 q^3\ge 0$$ $$\iff (p^2-3q)^3+2(p^2-3q)^2q+(q^2-3p)q \ge 0$$

which is obvious using $p^2\ge 3q$ and $q^2 \ge 3p$.

Answer 2

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$ Hence, we need to prove that $f(w^3)\geq0$, where $$f(w^3)=(9u^2-6v^2)^3-(27u^2-27uv^2+3w^3)9uv^2$$ and we see that $f$ is a decreasing function, which says

that it's enough to prove our inequality for a maximal value of $w^3$,

which happens for equality case of two variables.

Since our inequality is homogeneous, we can assume $b=c=1$, which gives $$(a-1)^2(a^4-2a+4)\geq0,$$ which is obviously true.

Done!

Answer 3

Here's a proof without using the $uvw$ method. (I'm late to the party, and now having to clean the mess. Also, credit goes to Michael Rozenberg for this solution.)

Continue to where you left off, we have that $$\frac{a^2 + b^2 + c^2}{ab + bc + ca} - 1\ge \frac{(a^3 + b^3 + c^3)(a + b + c)}{(a^2 + b^2 + c^2)^2} - 1$$

$$\iff \frac{\displaystyle \sum_{cyc}(a - b)^2}{2(ab + bc + ca)} \ge \frac{\displaystyle \sum_{cyc}ab(a - b)^2}{(a^2 + b^2 + c^2)^2}$$

Furthermore, because of $ab + bc + ca \le a^2 + b^2 + c^2$

$$\iff \frac{\displaystyle \sum_{cyc}ab(a - b)^2}{(a^2 + b^2 + c^2)^2} \le \frac{\displaystyle \sum_{cyc}ab(a - b)^2}{(ab + bc + ca)(a^2 + b^2 + c^2)}$$

Now, what needs to be proven is $$\sum_{cyc}(a - b)^2 \ge \frac{\displaystyle \sum_{cyc}ab(a - b)^2}{2(a^2 + b^2 + c^2)} \iff \sum_{cyc}\left(\frac{1}{2} - \frac{ab}{a^2 + b^2 + c^2}\right)(a - b)^2 \ge 0$$

$$\iff \sum_{cyc}\left[\frac{(a - b)^2 + c^2}{2(a^2 + b^2 + c^2)}\right](a - b)^2 \ge 0$$

In conclusion, the Sum of Square method does work.