Prove that $\sqrt{x^{3} + y^{3}+1} + \sqrt{z^{3} + y^{3}+1} + \sqrt{x^{3} + z^{3}+1} \ge 2 + \sqrt{2(x^{3} + y^{3}+z^{3})+1}$

Question

Given $x, y,z \ge 0$. Prove that $$\sqrt{x^{3} + y^{3}+1} + \sqrt{z^{3} + y^{3}+1} + \sqrt{x^{3} + z^{3}+1} \ge 2 + \sqrt{2(x^{3} + y^{3}+z^{3})+1} $$

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Attempt

Notice that $$(x^{3} + y^{3}+1) + (z^{3} + y^{3}+1) + (x^{3} + z^{3}+1) = 2 + 2(x^{3} + y^{3}+z^{3})+1 $$

So the sum of squares must be between two forms as below:

$$2 + \sqrt{2(x^{3} + y^{3}+z^{3})+1} \le \sqrt{x^{3} + y^{3}+1} + \sqrt{z^{3} + y^{3}+1} + \sqrt{x^{3} + z^{3}+1} \le 2 + 2(x^{3} + y^{3}+z^{3})+1$$

If I work backwards I will get, by letting the form in each root be $a,b,c$ respectively:

$$\sqrt{a} + \sqrt{b} + \sqrt{c} \ge 2 + \sqrt{a+b+c-2} $$

$$ a + b +c + 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) \ge 4 + (a+b+c-2) + 4\sqrt{a+b+c-2}$$

$$ (\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) \ge 1 + 2\sqrt{a+b+c-2}$$

$$ ((\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) -1))^{2} \ge 4(a+b+c-2)$$

$$ (ab + bc + ca + 2a^{2}\sqrt{bc} +2b^{2}\sqrt{ac} +2c^{2}\sqrt{ab}) - 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) + 1 \ge 4(a+b+c-2)$$

How to continue?

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Apart from the attempt above, I have tried:

• AM-GM for the sum of the 3 squares

• AM-GM for the sum of the 3 squares, then using Holder's inequality from the GM

• AM-GM for each square

Answer 1

The inequality $$\sum_{cyc}\sqrt{a+1}\geq2+\sqrt{a+b+c+1}$$ we can prove also by the following way: $$\sum_{cyc}\sqrt{a+1}-2-\sqrt{a+b+c+1}=$$ $$=\sqrt{a+1}+\sqrt{b+1}-2-\left(\sqrt{a+b+c+1}-\sqrt{c+1}\right)=$$ $$=\sqrt{a+1}+\sqrt{b+1}-2-\frac{a+b}{\sqrt{a+b+c+1}+\sqrt{c+1}}\geq$$ $$\geq\sqrt{a+1}+\sqrt{b+1}-2-\frac{a+b}{\sqrt{a+b+1}+1}=$$ $$=\sqrt{a+1}+\sqrt{b+1}-2-(\sqrt{a+b+1}-1)=$$ $$=\sqrt{a+1}-1-\left(\sqrt{a+b+1}-\sqrt{b+1}\right)=$$ $$=\sqrt{a+1}-1-\frac{a}{\sqrt{a+b+1}+\sqrt{b+1}}\geq\sqrt{a+1}-1-\frac{a}{\sqrt{a+1}+1}=0.$$

Answer 2

It should be $$ (ab + bc + ca + 2a\sqrt{bc} +2b\sqrt{ac} +2c\sqrt{ab}) - 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca}) + 1 \ge 4(a+b+c-2),$$ where $a\geq1$, $b\geq1$ and $c\geq1$, for which it's not so easy to find a proof.

By the way, the Karamata's inequality helps.

Indeed, let $f(x)=\sqrt{x+1},$ $x^3+y^3=c$, $x^3+z^3=b$ and $y^3+z^3=a$.

Thus, $f$ is a concave function and we need to prove that $$\sum_{cyc}\sqrt{a+1}\geq2+\sqrt{a+b+c+1},$$ which is just Karamata because if $a\geq b\geq c$ so $$(a+b+c,0,0)\succ(a,b,c)$$ and $$f(a+b+c)+f(0)+f(0)\leq f(a)+f(a)+f(b)$$ and we are done!