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

Question

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

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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 $$a,b,c \geq 0$$

Answer 1

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 $a+b+c=3$).

Answer 2

From the original inequality $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$

Let’s set $$x= \sqrt{3a + b^3}$$ $$y= \sqrt{3b + c^3}$$ $$z= \sqrt{3c + a^3}$$ By AM-GM we know that $$x^2-2xy+y^2 \ge 0$$ $$y^2-2yz+z^2 \ge0$$ $$z^2-2xz+x^2 \ge 0$$ So we then can set up the following inequalities

$$x^2+y^2\ge2xy$$ $$y^2+z^2\ge2yz$$ $$z^2+x^2\ge2xz$$

Multiply each inequality above by $z$, $x$, $y$ in that order which yields $$zx^2+zy^2\ge2xyz$$ $$xy^2+xz^2\ge2xyz$$ $$yz^2+yx^2\ge2xyz$$ Add these inequalities and factor to get $$xy(x+y)+xz(x+z)+zy(y+z) \ge 6xyz$$ Now divide both sides by $xyz$ for $$\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}\ge6$$ Now replace $x$, $y$ and $z$ with the terms on the LHS of the original inequality above $$\frac{\sqrt{3a + b^3}+\sqrt{3b + c^3}}{\sqrt{3c + a^3}}+\frac{\sqrt{3a + b^3}+\sqrt{3c + a^3}}{\sqrt{3b+ c^3}}+ \frac{\sqrt{3b + c^3}+\sqrt{3c + a^3}}{\sqrt{3a+ b^3}} \ge 6$$ with equality if and only if $a=b=c=1$.

Answer 3

I think your first method can still do something, but you probably have to use it in some other way, since using Cauchy-Schwarz directly will fail.

This is a suggestion.

Suppose

$\sqrt{3a+b^3} = u$

$\sqrt{3b+c^3} = v$

$\sqrt{3c+a^3} = w$

Then

$u^2+v^2+w^2 = 3(a+b+c)+a^3+b^3+c^3 = 9 + \sum a^3$

we are going to prove $u+v+w\ge 6$, which is equivalent to prove

$$u^2+v^2+w^2+2uv+2vw+2wu\ge 36$$

That means to prove

$$9 + \sum a^3+2uv+2vw+2wu\ge 36$$

we have proved

$u\ge \dfrac{1}{2}(3\sqrt{a}+\sqrt{b^3})$, thus

$2uv+2vw+2wu \ge \dfrac{1}{2}\sum(9\sqrt{ab}+\sqrt{b^3c^3}+3b^2+3\sqrt{ac^3}) = \\\dfrac{1}{2}\sum( 9\sqrt{ab} +\sqrt{a^3b^3}+3a^2 + 3\sqrt{ab^3})$

We then prove

$$\sum a^3 + \dfrac{1}{2}\sum( 9\sqrt{ab} +\sqrt{a^3b^3}+3a^2 + 3\sqrt{ab^3})\ge 27$$

I checked this with MATLAB, $(1,1,1)$ indeed is a local minimum, but I do not how to deal with the unsymmetric term $\sqrt{ab^3}$, I have tried to use $\sqrt{b^3}\ge 3\sqrt{b}-2$ to make it symmetric, but this will make the inequality fail.

Answer 4

I'm not sure if "proof with Mathematica" counts, but here it is:

In order to find the local extrema of function $f(a,b,c) = \sqrt{3a+b^3}+\sqrt{3b+c^3}+\sqrt{3c+a^3}$ under the constraint $a+b+c=3$, we differentiate the corresponding lagrangian and arrive at a system of equations

$$\left\{\begin{aligned}& (a^2c^2-a^2+1)^2(3a+b^3)=(b^2a^2-b^2+1)^2(3b+c^3)=(c^2b^2-c^2+1)^2(3c+a^3) \\& a+b+c=3 \\& c \ge a,b \ge 0 \end{aligned}\right.$$

Obviously, $a=b=c=1$ is one solution; with $f(a,b,c)=6$. In order to find the other solutions I plug this system into Mathematica. After putting the pieces together, Mathematica simplifies this to a 48-degree equation, with only 1 additional solution satisfying the constraints: $a \approx 0.335288, b \approx 0.744425, c \approx 1.920288$. The precision does not matter much, since at this point the value of the function is $f(a,b,c) \approx 6.65093$, which means it's a local maximum.

Answer 5

By Holder $$\left(\sum_{cyc}\sqrt{3a+b^3}\right)^2\sum_{cyc}\frac{(a+b)^3}{3a+b^3}\geq8(a+b+c)^3=216.$$ Thus, it's enough to prove that: $$\sum_{cyc}\frac{(a+b)^3}{3a+b^3}\leq6$$ or $$\sum_{cyc}\frac{(a+b)^3}{a(a+b+c)^2+3b^3}\leq2$$ or $$\sum_{cyc}(2a^8b+5a^7b^2-3a^6b^3+3a^6c^3-a^5b^4+17a^5c^4)+$$ $$+abc\sum_{cyc}(12a^6+20a^5b+11a^5c-10a^4b^2-11a^4c^2-3a^4b^3-19a^3b^2c-32a^3c^2b+9a^2b^2c^2)\geq0$$ or $$1.5\sum_{cyc}(a^8b+a^4b^5-2a^6b^3)+0.5\sum_{cyc}(a^7b^2+a^3b^6-2a^5b^4)+9abc\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)+$$ $$+3abc\sum_{cyc}(a^6-a^3b^3)+2abc\sum_{cyc}(a^5b+a^5c-a^4b^2-a^4c^2)+$$ $$+\sum_{cyc}(0.5a^8b+4.5a^7b^2+2.5a^6c^3+15.5a^5c^4+18a^6b^2c+9a^6c^2b+a^5b^3c-19a^4b^3c^2-32a^4c^3b^2)\geq0,$$ which is true by Schur, Muirhead and AM-GM.

Answer 6

Proof by KaiRain.

By squaring and collecting in terms, the original inequality is equivalent with: \begin{align*} \sum_{cyc} \sqrt{ \left(3a+b^3 \right) \left(3b+c^3 \right)} \ge \frac{3 \left(a+b \right) \left(b+c \right) \left(c+a \right)}{2} \end{align*} We will prove that: \begin{align} \sqrt{ \left(3a+b^3 \right) \left(3b+c^3 \right)} \ge \frac{3ab^2+b^2c}{2}+c^2a+abc \ \ \ (1)\end{align} holds true for all non-negative reals $a,b,c$ such that $a+b+c=3$. Indeed, this inequality can be proved by C-S and AM-GM in a rather tricky way. Following are the details: \begin{align*}3= \frac{ \left(a+b+c \right)^2}{3} = ab+bc+ca + \frac{ \left(b+c-2a \right)^2 +3\left(b-c \right)^2 }{12} \ge ab+bc+ca+ \frac{ \left(b-c \right)^2}{4} \end{align*} Using the above estimation, it turns out: \begin{align*} \sqrt{ \left(b^3+3a \right) \left(c^3+3b \right)} & \ge \sqrt{ b^3+a \left(ab+bc+ca \right) } \ \sqrt{c^3 + b \left(ab+bc+ca \right) +\frac{b \left(b-c \right)^2}{4}} \\ & = \sqrt{b^3+a^2b+ca^2+abc} \ \sqrt{ \frac{ b\left(b+c \right)^2}{4}+ab^2+c^3+abc} \\ & \ge \frac{b^2\left(b+c \right)}{2}+\sqrt{a^3b^3}+c^2a+abc\\ &= \frac{ b^3+\sqrt{a^3b^3}+\sqrt{a^3b^3} }{2} +\frac{b^2c}{2}+c^2a+abc\\ &\ge \frac{3ab^2}{2} +\frac{b^2c}{2}+c^2a+abc.\end{align*} Now, by adding two similar inequalities with $(1)$ and taking a note that: \begin{align*} \sum_{cyc} \left(\frac{3ab^2+b^2c}{2}+c^2a+abc \right) = \frac{3 \left(a+b \right) \left(b+c \right) \left(c+a \right)}{2} \end{align*} the result follows. Hence, the proof is completed.