Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$

If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that: $$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$

Here's what I've tried:

Using Cauchy-Schawrz I proved that:

$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$

$$\sqrt{(3a + b^3)(4)} \ge 3\sqrt{a} + \sqrt{b^3}$$

$$\sqrt{(3a + b^3)} \ge \frac{3\sqrt{a} + \sqrt{b^3}}{2}$$

Also I get:

$$\sqrt{(3b + c^3)} \ge \frac{3\sqrt{b} + \sqrt{c^3}}{2}$$

$$\sqrt{(3c + a^3)} \ge \frac{3\sqrt{c} + \sqrt{a^3}}{2}$$

If I add add 3 inequalities I get:

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

Now i need to prove that:

$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} + \frac{3\sqrt{b} + \sqrt{b^3}}{2} + \frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge 6 = 2(a+b+c)$$

It's enough now to prove that:

$$\frac{3\sqrt{a} + \sqrt{a^3}}{2} \ge b+c = 3-a$$ $$\frac{3\sqrt{b} + \sqrt{b^3}}{2} \ge a+c = 3-b$$ $$\frac{3\sqrt{c} + \sqrt{c^3}}{2} \ge b+a = 3-c$$

All three inequalities are of the form:

$$\frac{3\sqrt{x} + \sqrt{x^3}}{2} \ge 3-x$$

$$3\sqrt{x} + \sqrt{x^3} \ge 6-2x$$

$$(3\sqrt{x} + \sqrt{x^3})^2 \ge (6-2x)^2$$

$$9x + x^3 + 6x^2 \ge 36 - 24x + 4x^2$$

$$x^3 + 2x^2 + 33x - 36 \ge 0$$

$$(x-1)(x^2 + 3x + 33) \ge 0$$

Case 1:

$$(x-1) \ge 0 \ \ \ \ \text{ for any }\ x \geq 1$$

$$(x^2 + 3x + 33) \ge 0 \ \ \ \ \text{ for any x in R} $$

Case 2:

$$0 \ge (x-1) \ \ \ \ \text{ for any }\ 1 \geq x$$

$$0 \ge (x^2 + 3x + 33) \ \ \ \ \text{there are no solutions in R} $$

This proves that for $$x \geq 1$$ $$(x-1)(x^2 + 3x + 33) \ge 0$$ is true and so it is $$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$$, but a, b, c can be every non-negative number. I proved it's true for $$a,b,c \geq 1$$, but i can't for $$a,b,c \geq 0$$

Solution 1:

maybe this following solution is by Vasc?

since use Cauchy-Schwarz inequality,we have $$(a^3+3b)(a+3b)\ge (a^2+3b)^2$$ It suffices to show that $$\sum_{cyc}\dfrac{a^2+3b}{\sqrt{a+3b}}\ge 6$$ By Holder $$\left(\sum_{cyc}\dfrac{a^2+3b}{\sqrt{a+3b}}\right)^2[\sum_{cyc}(a^2+3b)(a+3b)]\ge[\sum_{cyc}(a^2+3b)]^3=[a^2+b^2+c^2+9]^3$$ it is enought to show that $$\left(a^2+b^2+c^2+9\right)^3\ge 36\sum_{cyc}(a^2+3b)(a+3b)$$ let $p=a+b+c=3,q=ab+bc+ac\le 3$,we have $$\sum_{cyc}(a^2+3b)(a+3b)=108-24q+3[abc+\sum_{cyc}a^2b]$$ use this well know：see inequality $$abc+\sum_{cyc}a^2b\le 4$$ we get $$\sum_{cyc}(a^2+3b)(a+3b)\le 24(5-q)$$ and $$a^2+b^2+c^2+9=2(9-q)$$ It suffices to show that $$(9-q)^3\ge 108(5-q)$$ which is true,because equivalent $$(q-3)^2(21-q)\ge 0$$

Solution 2:

Cauchy-Schwarz inequality is not the way to go here. Notice that at $a = b = c =1$ your original inequality becomes exact, whereas your relaxed inequality is no longer satisfied. Moreover, you have lost a factor of 2 there.

Your new edits are very interesting, but still lead to a dead-end: you can check that when $a=b=0$, $c=3$ your inequality labeled "Now I need to prove that:" fails. Also note that in general the case $a,b,c\ge 1$ contains only one point $a = b = c = 1$ in it (because $a+b+c=3$).