Proving two root inequalities using AM GM and averages

Expert intro: The inequalities $\sqrt{xy} \le \sqrt{\frac{x^2+y^2}{2}}$ and $\sqrt[4]{abcd} \le \sqrt{\frac{a^2+b^2+c^2+d^2}{4}}$ both come from applying AM-GM to squared variables. The given estimate for positive real numbers is the bridge that turns a geometric mean into a power mean comparison [1].

Detailed walkthrough

Part (i): from $\sqrt{xy}$ to a root of squares

For positive real numbers $x$ and $y$, we are given that $\sqrt{xy} \le \frac{x+y}{2}.$ To prove $\sqrt{xy} \le \sqrt{\frac{x^2+y^2}{2}},$ apply the same AM-GM idea to the positive numbers $x^2$ and $y^2$: $\sqrt{x^2y^2} \le \frac{x^2+y^2}{2}.$ Since $x,y>0$, we have $\sqrt{x^2y^2}=xy$. Taking square roots of both sides gives $\sqrt{xy} \le \sqrt{\frac{x^2+y^2}{2}}.$ That is the required result.

The key observation is that the inequality compares a geometric mean with a quadratic average. Squaring the variables first makes the given AM-GM inequality fit the target exactly.

Part (ii): extending to four variables

For positive real numbers $a,b,c,d$, we want to show $\sqrt[4]{abcd} \le \sqrt{\frac{a^2+b^2+c^2+d^2}{4}}.$ Apply AM-GM to the four positive numbers $a^2,b^2,c^2,d^2$: $\sqrt[4]{a^2b^2c^2d^2} \le \frac{a^2+b^2+c^2+d^2}{4}.$ Because all variables are positive, $\sqrt[4]{a^2b^2c^2d^2}=\sqrt{abcd}.$ Now take square roots of both sides: $\sqrt[4]{abcd} \le \sqrt{\frac{a^2+b^2+c^2+d^2}{4}}.$ This is exactly the desired inequality.

The structure is the same as in part (i): first compare geometric and arithmetic means of squared quantities, then take the appropriate root to match the original expression.

Why this method works

The expression on the right is an RMS-type quantity, because $\sqrt{\frac{x^2+y^2}{2}}$ is the root mean square of $x$ and $y$. The root mean square is always at least the geometric mean. Likewise, for four values, $\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}$ is at least the fourth root of the product.

That is why both inequalities are true: they are instances of the general fact that the arithmetic mean of squares dominates the square of the arithmetic mean, and hence dominates the corresponding geometric mean after taking square roots.

Common mistake to avoid

A common error is to try to compare $\sqrt{xy}$ directly with $\frac{x^2+y^2}{2}$ without checking the exponents. The target has a square root on the right, so the proof must keep track of which side is being rooted and when. Another mistake is to use AM-GM on $x$ and $y$ again in part (i) and assume the result immediately matches the claim. It does not: you need AM-GM on $x^2$ and $y^2$. In part (ii), the same care is needed with the fourth root. Every exponent must line up before taking roots [2].

💡 Pitfall guide

The step that usually causes trouble in $\sqrt{xy} \le \sqrt{\frac{x^2+y^2}{2}}$ is forgetting that the given AM-GM statement applies to positive inputs, so you should feed it $x^2$ and $y^2$, not $x$ and $y$ directly. If you apply AM-GM to $x$ and $y$, you get $\sqrt{xy} \le \frac{x+y}{2}$, which is true but not the target. Another subtle issue is taking square roots before the inequality has been written in the correct form. You must first obtain $xy \le \frac{x^2+y^2}{2}$, then take square roots, because both sides are positive. In the four-variable part, students often misread $\sqrt[4]{abcd}$ as something like $\sqrt{abcd}$, which changes the exponent count and breaks the argument. Keep the exponents aligned all the way through.

🔄 Real-world variant

If the first inequality were changed to prove $\sqrt{xy} \le \frac{x+y}{2}$ for positive real numbers $x$ and $y$, then the given AM-GM statement already solves it directly. If the second inequality were changed to $\sqrt[4]{abcd} \le \frac{a+b+c+d}{4}$, you would apply AM-GM once to the four numbers $a,b,c,d$ and obtain the result immediately. A slightly different variant, such as proving $\sqrt[3]{pqr} \le \sqrt{\frac{p^2+q^2+r^2}{3}}$, would need a new idea because the exponent pattern no longer matches a simple square-root form. The original question works smoothly because squaring the variables converts the target into a direct AM-GM application.

🔍 Related terms

arithmetic mean geometric mean, root mean square, power mean inequality