如何证明当x²+y²+z²=1且x,y,z>0时，x²yz+xy²z+xyz²≤1/3？

Alright, let's break down how to prove this inequality step by step. We need to show that for positive real numbers (x, y, z) satisfying (x^2 + y^2 + z^2 = 1), the inequality (x^2yz + xy^2z + xyz^2 \leq \frac{1}{3}) holds.

First, let's simplify the left-hand side (LHS) to make it easier to work with:
$$x^2yz + xy^2z + xyz^2 = xyz(x + y + z)$$
Our goal now is to bound this simplified expression using the given constraint (x^2 + y^2 + z^2 = 1).

Step 1: Bound (xyz) using the AM-GM Inequality #

For any positive real numbers (a, b, c), the Arithmetic Mean-Geometric Mean (AM-GM) inequality states:
$$\frac{a + b + c}{3} \geq \sqrt[3]{abc}$$
Let (a = x^2), (b = y^2), (c = z^2). Since we know (x^2 + y^2 + z^2 = 1), substitute into the AM-GM formula:
$$\frac{1}{3} \geq \sqrt[3]{x^2y2z^2}$$
Cube both sides to eliminate the cube root:
$$\left(\frac{1}{3}\right)^3 \geq x^2y2z^2 \implies \frac{1}{27} \geq x^2y2z^2$$
Take the square root of both sides (since (x, y, z) are positive, we don't have to worry about negative values):
$$xyz \leq \frac{1}{3\sqrt{3}}$$

Step 2: Bound (x + y + z) using the Cauchy-Schwarz Inequality #

The Cauchy-Schwarz inequality tells us that for real numbers (p_1, p_2, p_3) and (q_1, q_2, q_3):
$$(p_1q_1 + p_2q_2 + p_3q_3)^2 \leq (p_1^2 + p_2^2 + p_3^2)(q_12 + q_2^2 + q_3^2)$$
Let (p_1 = p_2 = p_3 = 1) and (q_1 = x), (q_2 = y), (q_3 = z). Applying Cauchy-Schwarz here:
$$(x + y + z)^2 \leq (1^2 + 1^2 + 1^2)(x2 + y^2 + z^2) = 3 \times 1 = 3$$
Take the square root of both sides:
$$x + y + z \leq \sqrt{3}$$

Step 3: Combine the Bounds #

Multiply the two inequalities we derived to get a bound on our simplified LHS:
$$xyz(x + y + z) \leq \frac{1}{3\sqrt{3}} \times \sqrt{3}$$
Simplify the right-hand side (RHS):
$$\frac{1}{3\sqrt{3}} \times \sqrt{3} = \frac{1}{3}$$
This gives us exactly the result we need:
$$x^2yz + xy^2z + xyz^2 \leq \frac{1}{3}$$

Step 4: Verify the Equality Condition #

Equality in both AM-GM and Cauchy-Schwarz holds when (x = y = z). Substitute this into our constraint (x^2 + y^2 + z^2 = 1):
$$3x^2 = 1 \implies x = y = z = \frac{1}{\sqrt{3}}$$
Plugging these values back into the LHS confirms it equals (\frac{1}{3}), so equality is indeed achievable.

内容的提问来源于stack exchange，提问作者hesham3886