If $x, y, z\in\mathbb R^+ $ and $x^3+y^3=z^3,$ then prove that $x^2+y^2-z^2>6(z-x) (z-y). $

Question

I made several unsuccessful attempts. Still couldn't think of a proper way to prove the inequality. Please suggest how to approach this problem. Thanks in advance.

EDIT 1. My approach (that I was talking about):

Given: $z^3=x^3+y^3.$

We have to prove:

$x^2+y^2-z^2>6(z-x) (z-y)$ i.e., $\underbrace{(z^2+zx+x^2) (z^2+zy+y^2) (x^2+y^2-z^2)}_{=E\text{ (say)}}>6(z^3-x^3) (z^3-y^3)=6x^3y^3.$

(Here one thing which I noticed is that $(x^2+y^2-z^2)>0,$ since each of the terms on the LHS except this one is positive and $6x^3y^3$ is also positive for $x, y, z>0.$)

Using AM $\ge$ GM, we have:

$E\ge 3zx\cdot3zy\cdot(x^2+y^2-z^2)\ge 9xyz^2(2xy-z^2).$

From here I couldn't think of a proper way to prove $E>6x^3y^3.$ But I'm still working on it. At present I'm trying to manipulate the expression $9xyz^2(2xy-z^2)$ to get the job done. If I find something useful I'll update here.

Answer 1

Hint: eliminate $z$ from the system $$ \eqalign{x^3 + y^3 &= z^3 \cr x^2 + y^2 - z^2 &= 6 (z-x)(z-y)\cr} $$ and show the only real solutions of the resulting equation are $x=0$ and $y=0$.

Answer 2

Let $x^2+y^2=2uxy$.

Thus, $u\geq1$ and we need to prove that: $$x^2+y^2-\sqrt[3]{(x^3+y^3)^2}>6\left(\sqrt[3]{x^3+y^3}-x\right)\left(\sqrt[3]{x^3+y^3}-y\right)$$ or $$x^2+y^2-6xy+6(x+y)\sqrt[3]{x^3+y^3}-7\sqrt[3]{(x^3+y^3)^2}>0$$ or $$(x^2+y^2-6xy)^3+216(x+y)^3(x^3+y^3)-343(x^3+y^3)^2+$$ $$+126(x^2+y^2-6xy)(x+y)(x^3+y^3)>0$$ or $$4(u-3)^3+432(u+1)^2(2u-1)-343(u+1)(2u-1)^2+$$ $$+252(u-3)(u+1)(2u-1)>0$$ or $$129u-127>0,$$ which is obvious.

Answer 3

Th proof by Bjkjdz. The inequality equivalent to $${x^2}z + {y^2}z - {z^3} > 6z(z - x)(z - y),$$ or $${x^2}(z - x) + {y^2}(z - y) > 6z(z - x)(z - y),$$ or $$\dfrac{{{x^2}}}{{z - y}} + \dfrac{{{y^2}}}{{z - x}} > 6z.$$ But $$\left\{ \begin{array}{l} {x^2} = \dfrac{{{x^3}}}{x} = \dfrac{{{z^3} - {y^3}}}{x} = \dfrac{{(z - y)({z^2} + yz + {y^2})}}{x} \\ {y^2} = \dfrac{{{y^3}}}{y} = \dfrac{{{z^3} - {x^3}}}{y} = \dfrac{{(z - x)({z^2} + zx + {x^2})}}{y} \\ \end{array} \right.$$ Thefore, we will show that $$\dfrac{{{y^2} + yz+ {z^2}}}{x} + \dfrac{{{z^2} + zx + {x^2}}}{y} > 6z,$$ equivalent to $$\left( {{x^3} + y{z^2} + {y^2}z} \right) + \left( {{y^3} + {x^2}z + x{z^2}} \right) > 6xyz.$$ Which is true by the AM-GM inequality.

Answer 4

Because each expression is homogenous, we may assume that $ z = 1$ (by using the substitution $ x' = \frac{x}{z}$). The question becomes:

If $x^3 + y^3 = 1$, show that $x^2 - 6xy + y^2 + 6x + 6 y - 7 > 0$.

Note: This 2-variable inequality type is common, and there are several ways of dealing with it by exploiting $x+y$.

Proof: Let $ w = x+y$. Since $ (x^3+y^3) < (x+y)^3 \leq 4(x^3 + y^3)$, thus $ 1 < w \leq \sqrt[3]{4} $.

$(x+y)(x^2 - 6xy + y^2 + 6x + 6 y - 7) \\ = x^3 +y^3 - 5xy (x+y) + 6(x+y)^2 - 7 (x+y) \\ = 1 - \frac{5}{3} ( (x+y)^3 - 1) + 6(x+y)^2 - 7 (x+y) \\ = \frac{1}{3}(-5 w^3 + 18w^2 - 21w + 8) \\ = \frac{1}{3}(w-1)^2(-5w+8) $

Since $ 1 < w \leq \sqrt[3]{4} < \frac{8}{5}$, the final expression is positive, and we conclude that $x^2 - 6xy + y^2 + 6x + 6 y - 7 > 0$.

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Of course, we didn't need to do the initial substitution, and could have shown that $$(x+y)\left[ x^2+y^2-z^2 - 6 ( z-y)(z-x)\right] = \frac{1}{3} ( x+y -z )^2 ( 8z - 5x - 5y )> 0.$$

However, it's hard to see how the equality holds by applying $(x+y)^3 = z^3 + xy(x+y)$.

Answer 5

The solution is due to @Calvin Lin.

Problem: Find the best constant $C$ such that $x^2+y^2 - 1 \ge C(1-x)(1-y)$ holds for all $x, y \ge 0$ with $x^3+y^3 = 1$.

Solution: The best constant is $C = 2^{4/3} + 2^{2/3} + 2$. Let us prove it.

Let $w = x + y$. Since $x^3+y^3 \le (x+y)^3 \le 4(x^3 + y^3)$, we have $1\le w \le \sqrt[3]{4}$. We have \begin{align} &(x+y)[x^2+y^2 - 1 - C(1-x)(1-y)]\\ =\ & x^3+y^3 - (C-1)xy(x+y) + C(x+y)^2 - (C+1)(x+y)\\ =\ & x^3+y^3 - \frac{C-1}{3}[(x+y)^3-x^3-y^3] + C(x+y)^2 - (C+1)(x+y)\\ =\ & 1 - \frac{C-1}{3}[(x+y)^3-1] + C(x+y)^2 - (C+1)(x+y)\\ =\ & \frac{1}{3}(w-1)^2[2 + w - C(w-1)]. \end{align} From $2 + w - C(w-1) \ge 0$ for $1\le w \le \sqrt[3]{4}$, we have $$C \le \inf_{1 < w \le \sqrt[3]{4}} \frac{2+w}{w-1} = 2^{4/3} + 2^{2/3} + 2.$$ On the other hand, when $C = 2^{4/3} + 2^{2/3} + 2$, we have $$2 + w - C(w-1) = (2^{4/3} + 2^{2/3} + 1)(\sqrt[3]{4} - w) \ge 0$$ for $1\le w \le \sqrt[3]{4}$. We are done.