# Prove that $\frac {x^3}{u}+\frac {y^3}{v} + \frac {z^3}{w}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x^2+y^2+z^2)$

Let $$x, y, z\in [a,b]$$ and $$u, v, w\in [a, b]$$, where $$0 , s.t. $$x^2+y^2+z^2=u^2+v^2+w^2$$. Show that $$\frac {x^3}{u}+\frac {y^3}{v} + \frac {z^3}{w}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x^2+y^2+z^2) .$$

I tried to apply Cauchy- Schwartz and the fact that $$\frac {a^4+b^4}{ab (a^2+b^2)} \geq \frac {a}{b}$$ but I didn't succeed.

• What about the rearrangement inequality? Sep 14, 2018 at 4:22
• @abc with rearrangement inequality I change the form of the inequality. You want to amplify each fraction and after that to apply the inequality?
– rafa
Sep 14, 2018 at 4:58

The inequality is homogeneous, i.e. multiplying all variables with the same factor will not change it. This allows us to demand that the multiplied variables (for which we use the same variable letters as before) satisfy $$x^2 + y^2 + z^2 = u^2 + v^2 + w^2 = 3$$ and in turn $$a \le 1$$ and $$\sqrt{3-a^2} \ge b \ge 1$$. So we have to prove

$$\frac {x^3}{u}+\frac {y^3}{v} + \frac {z^3}{w}\leq 3 \frac {a^4+b^4}{ab (a^2+b^2)}$$

By re-ordering, w.l.o.g., let $$x \le y\le z$$. By the rearrangement inequality, the LHS gets as large as possible if we use, for $$z^3$$, the smallest possible $$w$$, which is $$a$$. Considering the choices of $$x,y,z$$ the highest value on the LHS will then be obtained for $$z=b$$. So we have to prove

$$\frac {x^3}{u}+\frac {y^3}{v} + \frac {b^3}{a}\leq 3 \frac {a^4+b^4}{ab (a^2+b^2)}$$

Again by rearrangement, the converse is true for the term $$\frac {x^3}{u}$$, so we need to prove

$$\frac {a^3}{b}+\frac {y^3}{v} + \frac {b^3}{a}\leq 3 \frac {a^4+b^4}{ab (a^2+b^2)}$$

Now the remaining values are determined: $$y^2=v^2=3-a^2-b^2$$. So finally we need to show

$$\frac {a^3}{b}+(3-a^2-b^2) + \frac {b^3}{a}\leq 3 \frac {a^4+b^4}{ab (a^2+b^2)}$$

which is

$$(a^2+b^2)(a^4 + ab(3-a^2-b^2) + b^4) \leq 3 (a^4+b^4)$$

Writing $$3=a^2+b^2+v^2$$ we have

$$(a^2+b^2)(a^4 + ab v^2 + b^4) \leq (a^2+b^2+v^2) (a^4+b^4)\\ \rightarrow \quad 0 \leq v^2 (b^3-a^3)(b-a)$$ which is true. This proves the claim.

Note that for more than three terms, a similar inequality holds. See here.

• I think we can not assume that $x\leq y\leq z$. Sep 29, 2018 at 10:23
• @MichaelRozenberg Re-ordering also changes the roles of $u,v,w$, which are discussed later. Sep 29, 2018 at 12:25

Proof:

First, we give some auxiliary results (Facts 1-4).

Fact 1: Let $$Q \in [3a^2, 3b^2]$$. Let $$X, Y, Z \in [a, b]$$ with $$X^2 + Y^2 + Z^2 = Q$$. Then $$X^3 + Y^3 + Z^3 \le \left\{\begin{array}{ll} (\sqrt{Q - 2a^2})^3 + a^3 + a^3 & \mathrm{if} ~ 3a^2 \le Q \le 2a^2 + b^2 \\[10pt] b^3 + (\sqrt{Q - a^2 - b^2})^3 + a^3 & \mathrm{if} ~ 2a^2 + b^2 < Q < a^2 + 2b^2\\[10pt] b^3 + b^3 + (\sqrt{Q- 2b^2})^3 & \mathrm{if} ~ a^2 + 2b^2 \le Q \le 3b^2. \end{array} \right.$$ (Proof: Note that $$t \mapsto t^{3/2}$$ is convex on $$t \ge 0$$. Also, $$(X^2, Y^2, Z^2)$$ is majorized by $$(Q - 2a^2, a^2, a^2)$$ if $$3a^2 \le Q \le 2a^2 + b^2$$, by $$(b^2, Q - a^2 - b^2, a^2)$$ if $$2a^2 + b^2 < Q < a^2 + 2b^2$$, and by $$(b^2, b^2, Q - 2b^2)$$ if $$a^2 + 2b^2 \le Q \le 3b^2$$. By Karamata's inequality, the desired results follow.)

Fact 2: Let $$Q \in [3a^2, 3b^2]$$. Let $$U, V, W\in [a, b]$$ with $$U^2 + V^2 + W^2 = Q$$. Then $$\frac{1}{U} + \frac{1}{V} + \frac{1}{W} \le \left\{\begin{array}{ll} \frac{1}{a} + \frac{1}{a} + \frac{1}{\sqrt{Q-2a^2}} & \mathrm{if} ~ 3a^2 \le Q \le 2a^2 + b^2 \\[10pt] \frac{1}{a} + \frac{1}{\sqrt{Q - a^2 - b^2}} + \frac{1}{b} & \mathrm{if} ~ 2a^2 + b^2 < Q < a^2 + 2b^2\\[10pt] \frac{1}{\sqrt{Q - 2b^2}} + \frac{1}{b} + \frac{1}{b} & \mathrm{if} ~ a^2 + 2b^2 \le Q \le 3b^2. \end{array} \right.$$ (Proof: Note that $$t\mapsto \frac{1}{\sqrt t}$$ is convex on $$t > 0$$. Also, $$(U^2, V^2, W^2)$$ is majorized by $$(Q - 2a^2, a^2, a^2)$$ if $$3a^2 \le Q \le 2a^2 + b^2$$, by $$(b^2, Q - a^2 - b^2, a^2)$$ if $$2a^2 + b^2 < Q < a^2 + 2b^2$$, and by $$(b^2, b^2, Q - 2b^2)$$ if $$a^2 + 2b^2 \le Q \le 3b^2$$. By Karamata's inequality, the desired results follow.)

Fact 3: Let $$p_1 \ge p_2 \ge p_3 > 0$$. Let $$Q \in [3a^2, 3b^2]$$. Let $$X, Y, Z \in [a, b]$$ with $$X^2 + Y^2 + Z^2 = Q$$. Then $$p_1 X^3 + p_2 Y^3 + p_3 Z^3 \le \left\{\begin{array}{ll} p_1 (\sqrt{Q - 2a^2})^3 + p_2 a^3 + p_3a^3 & \mathrm{if} ~ 3a^2 \le Q \le 2a^2 + b^2 \\[10pt] p_1b^3 + p_2(\sqrt{Q - a^2 - b^2})^3 + p_3a^3& \mathrm{if} ~ 2a^2 + b^2 < Q < a^2 + 2b^2\\[10pt] p_1 b^3 + p_2 b^3 + p_3(\sqrt{Q - 2b^3})^3& \mathrm{if} ~ a^2 + 2b^2 \le Q \le 3b^2. \end{array} \right.$$ (Proof: It is easy to prove by using Fact 1. For example, if $$3a^2 \le Q \le 2a^2 + b^2$$, using Fact 1, we have $$\mathrm{RHS} - \mathrm{LHS} = p_1((\sqrt{Q-2a^2})^3 - X^3) - p_2(Y^3 - a^3) - p_3(Z^3 - a^3)$$ $$\ge p_1((\sqrt{Q-2a^2})^3 - X^3) - p_1(Y^3 - a^3) - p_1(Z^3 - a^3)$$ $$= p_1((\sqrt{Q-2a^2})^3 + 2a^2 - X^3 - Y^3 - Z^3) \ge 0$$.)

Fact 4: Let $$q_1 \ge q_2 \ge q_3 > 0$$. Let $$Q \in [3a^2, 3b^2]$$. Let $$U, V, W\in [a, b]$$ with $$U^2 + V^2 + W^2 = Q$$. Then $$\frac{q_1}{U} + \frac{q_2}{V} + \frac{q_3}{W} \le \left\{\begin{array}{ll} \frac{q_1}{a} + \frac{q_2}{a} + \frac{q_3}{\sqrt{Q-2a^2}} & \mathrm{if} ~ 3a^2 \le Q \le 2a^2 + b^2 \\[10pt] \frac{q_1}{a} + \frac{q_2}{\sqrt{Q - a^2 - b^2}} + \frac{q_3}{b} & \mathrm{if} ~ 2a^2 + b^2 < Q < a^2 + 2b^2\\[10pt] \frac{q_1}{\sqrt{Q - 2b^2}} + \frac{q_2}{b} + \frac{q_3}{b} & \mathrm{if} ~ a^2 + 2b^2 \le Q \le 3b^2. \end{array} \right.$$ (Proof: It is easy to prove by using Fact 2.)

Now let us proceed.

WLOG, assume that $$x \ge y \ge z$$ and $$u \le v \le w$$ (See Remarks at the end).

Let $$S = x^2 + y^2 + z^2 = u^2 + v^2 +w^2.$$ Clearly, $$3a^2 \le S \le 3b^2$$.

We split into three cases:

Case 1: $$3a^2 \le S \le 2a^2 + b^2$$

We have \begin{align*} \frac {x^3}{u}+\frac {y^3}{v} + \frac {z^3}{w} &\le \frac {x^3}{a}+\frac {y^3}{a} + \frac {z^3}{\sqrt{S - 2a^2}} \tag{1}\\ &\le \frac {(\sqrt{S- 2a^2})^3}{a}+\frac {a^3}{a} + \frac {a^3}{\sqrt{S - 2a^2}}\tag{2}\\ &\le \frac {a^4+b^4}{ab (a^2+b^2)} S. \tag{3} \end{align*} Explanations:
(1): Fact 2.
(2): Fact 1.
(3): Hint: Let $$S = 2a^2 + t^2$$ where $$t\in [a, b]$$.

Case 2: $$a^2 + 2b^2 \le S \le 3b^2$$

We have \begin{align*} \frac {x^3}{u}+\frac {y^3}{v} + \frac {z^3}{w} &\le \frac {x^3}{\sqrt{S - 2b^2}}+\frac {y^3}{b} + \frac {z^3}{b} \tag{4}\\ &\le \frac {b^3}{\sqrt{S - 2b^2}}+\frac {b^3}{b} + \frac {(\sqrt{S - 2b^2})^3}{b} \tag{5}\\ &\le \frac {a^4+b^4}{ab (a^2+b^2)} S. \tag{6} \end{align*} Explanations:
(4): Fact 2.
(5): Fact 1.
(6): Hint: Let $$S = 2b^2 + t^2$$ where $$t\in [a, b]$$.

Case 3: $$2a^2 + b^2 < S < a^2 + 2b^2$$

We have \begin{align*} \frac {x^3}{u}+\frac {y^3}{v} + \frac {z^3}{w} &\le \frac {x^3}{a}+\frac {y^3}{\sqrt{S - a^2 - b^2}} + \frac {z^3}{b}\tag{7}\\ &\le \frac {b^3}{a}+\frac {(\sqrt{S - a^2 - b^2})^3}{\sqrt{S - a^2 - b^2}} + \frac {a^3}{b} \tag{8}\\ &\le \frac {a^4+b^4}{ab (a^2+b^2)} S. \tag{9} \end{align*} Explanations:
(7): Fact 2.
(8): Fact 1.
(9): Hint: Let $$S = a^2 + b^2 + t^2$$ where $$t\in (a, b)$$.

We are done.

Remarks: Let $$u_1 \le v_1 \le w_1$$ be the rearrangement of $$u, v, w$$. Let $$x_1 \ge y_1 \ge z_1$$ be the rearrangement of $$x, y, z$$. By the rearrangement inequality, we have $$\frac {x^3}{u}+\frac {y^3}{v} + \frac {z^3}{w} \le \frac {x_1^3}{u_1}+\frac {y_1^3}{v_1} + \frac {z_1^3}{w_1}.$$