Proof Methods for Inequalities

Transform the inequality step-by-step using equivalent operations until you reach a statement that is obviously true (or obviously false). Key rules:

• Adding the same value to both sides preserves the inequality.
• Multiplying both sides by a positive constant preserves the inequality.
• Multiplying both sides by a negative constant reverses the inequality.
• If $a > b > 0$ and $c > d > 0$, then $ac > bd$.

Prove that $\dfrac{a^2 + b^2}{2} \geq ab$ for all real numbers $a, b$.

Proof. Consider the difference:

$a^2 + b^2 - 2ab = (a - b)^2 \geq 0$

Since $(a - b)^2$ is a perfect square, it is always non-negative. Therefore $a^2 + b^2 \geq 2ab$, and dividing both sides by $2$ gives:

$\frac{a^2 + b^2}{2} \geq ab \quad \square$

To prove $A \geq B$, show that $A - B$ can be written as a sum of squares or another expression that is clearly $\geq 0$.

Prove that $a^4 + b^4 \geq a^2 b^2 + a^2 b^2$ for all real $a, b$.

Proof. Compute the difference:

$a^4 + b^4 - 2a^2 b^2 = (a^2 - b^2)^2 \geq 0$

Since $(a^2 - b^2)^2 \geq 0$, we have $a^4 + b^4 \geq 2a^2 b^2$. $\square$

To prove statement $P$, assume $\neg P$ (the opposite). Derive a logical contradiction. Since the assumption leads to an impossibility, $P$ must be true.

Prove that if $a + b > 2$, then $a > 1$ or $b > 1$.

Proof. Assume for contradiction that both $a \leq 1$ and $b \leq 1$. Then:

$a + b \leq 1 + 1 = 2$

This contradicts the hypothesis $a + b > 2$. Therefore at least one of $a, b$ must be greater than $1$. $\square$

Divide the problem into exhaustive cases. Prove the inequality holds in each case separately. The union of all cases covers every possibility.

Prove that $|a + b| \leq |a| + |b|$ for all real $a, b$.

Proof. We consider cases based on the signs of $a$ and $b$.

Case 1: $a \geq 0$ and $b \geq 0$. Then $|a+b| = a+b = |a|+|b|$. The inequality holds with equality.

Case 2: $a < 0$ and $b < 0$. Then $|a+b| = -(a+b) = (-a)+(-b) = |a|+|b|$. Equality again.

Case 3: Mixed signs (say $a \geq 0$, $b < 0$; the other sub-case is symmetric). Then $|a+b| \leq \max(|a|, |b|) \leq |a| + |b|$, since the remaining term is non-negative.

In every case, $|a + b| \leq |a| + |b|$. $\square$

Prove that $\sqrt{ab} \leq \dfrac{a + b}{2}$ for all $a, b \geq 0$.

Hint: Start by squaring both sides (valid since both sides are non-negative) and rearrange.