The AM-GM and Cauchy-Schwarz Inequalities

Problem Statement

In Extension 2, proving inequalities rarely involves just shifting terms across a greater-than sign. Instead, we rely on established, heavily-tested bounding theorems. The two pillars of inequality proofs are the AM-GM Inequality and the Cauchy-Schwarz Inequality.

1. AM-GM: The Arithmetic Mean of non-negative real numbers is always greater than or equal to their Geometric Mean. For two numbers $a, b \ge 0$: $\frac{a+b}{2} \ge \sqrt{ab}$. (Equality holds when $a=b$).
2. Cauchy-Schwarz: For any real sequences $(u_1, u_2...)$ and $(v_1, v_2...)$: $(u_1v_1 + u_2v_2 + \dots)^2 \le (u_1^2 + u_2^2 + \dots)(v_1^2 + v_2^2 + \dots)$.

AM-GM is perfect for relating sums and products, while Cauchy-Schwarz is unparalleled for relating sums of products to sums of squares.

Consider three positive real numbers $x, y, z$.

(a) Use the AM-GM inequality on the pairs $(x, y)$, $(y, z)$, and $(z, x)$ to write down three separate inequalities.

(b) Multiply the three inequalities from part (a) together to prove that $(x+y)(y+z)(z+x) \ge 8xyz$.

(c) Given that $a^2 + b^2 = 1$ and $c^2 + d^2 = 1$ (where $a,b,c,d$ are real numbers), use the Cauchy-Schwarz inequality to find the maximum possible value of $ac + bd$.

Hints

• Part (a): Apply the formula $\frac{a+b}{2} \ge \sqrt{ab}$ directly to the given pairs. Multiply both sides by 2 to get it in the form $(a+b) \ge 2\sqrt{ab}$.
• Part (b): Since $x, y, z$ are positive, all terms in your three inequalities are positive. This means you can safely multiply the Left Hand Sides together and the Right Hand Sides together without flipping the inequality sign.
• Part (c): Write down the 2-variable algebraic form of Cauchy-Schwarz: $(u_1v_1 + u_2v_2)^2 \le (u_1^2 + u_2^2)(v_1^2 + v_2^2)$. Substitute $u_1=a$, $u_2=b$ and $v_1=c$, $v_2=d$. Use the given constraints to simplify the right-hand side.

Solutions

Part (a): Applying AM-GM

1. Applying AM-GM to $x$ and $y$: $\frac{x+y}{2} \ge \sqrt{xy} \implies (x+y) \ge 2\sqrt{xy}$
2. Applying AM-GM to $y$ and $z$: $\frac{y+z}{2} \ge \sqrt{yz} \implies (y+z) \ge 2\sqrt{yz}$
3. Applying AM-GM to $z$ and $x$: $\frac{z+x}{2} \ge \sqrt{zx} \implies (z+x) \ge 2\sqrt{zx}$

Part (b): Combining AM-GM Inequalities

1. Because $x, y, z > 0$, all sides of the inequalities are positive. We can multiply them together: $(x+y)(y+z)(z+x) \ge (2\sqrt{xy}) \times (2\sqrt{yz}) \times (2\sqrt{zx})$
2. Simplify the right hand side: $(x+y)(y+z)(z+x) \ge 8 \times \sqrt{xy \cdot yz \cdot zx}$ $(x+y)(y+z)(z+x) \ge 8 \times \sqrt{x^2 y^2 z^2}$
3. Since $x, y, z$ are positive, $\sqrt{x^2 y^2 z^2} = xyz$: $(x+y)(y+z)(z+x) \ge 8xyz$
4. (Note: Equality holds if and only if $x=y$, $y=z$, and $z=x$, which implies $x=y=z$.)

Part (c): Applying Cauchy-Schwarz

1. The Cauchy-Schwarz inequality for two variables is: $(ac + bd)^2 \le (a^2 + b^2)(c^2 + d^2)$
2. We are given the constraints $a^2 + b^2 = 1$ and $c^2 + d^2 = 1$. Substitute these into the right side: $(ac + bd)^2 \le (1)(1)$ $(ac + bd)^2 \le 1$
3. Taking the square root of both sides gives the absolute value boundary: $-1 \le ac + bd \le 1$
4. Therefore, the maximum possible value of the expression $ac + bd$ is 1.
5. (Geometrically, this represents the dot product of two unit vectors, which reaches its maximum of 1 when the vectors are perfectly aligned).

Takeaways

• The Fundamental Building Blocks: Most 3 or 4-mark inequality proofs in the HSC are just clever applications of AM-GM or Cauchy-Schwarz in disguise.
• The Power of Multiplication: If you have multiple inequalities where all terms are strictly positive, multiplying them together is a highly effective way to create complex new inequalities.
• Look for the Squares: If a question asks you to maximize a sum of products (like $ac+bd$) and gives you sums of squares (like $a^2+b^2$), that is the universal signal to use the Cauchy-Schwarz inequality.

Further Readings

• HSC Inequalities: https://vumaths.com/booklets/hsc-inequalities/
• HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
• HSC Vectors: https://vumaths.com/booklets/hsc-vectors/

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Inequalities are often considered the hardest topic in Extension 2 because there is no single algorithmic way to solve them. You need intuition! To build that intuition, download my comprehensive inequality guide on Vu's Maths Hub. Watch how I dissect hard trial questions on my YouTube channel. Follow my Instagram for daily maths tips, and subscribe to my Substack for deeper theoretical explorations!