Inequalities & Bounds

An inequality is useful when it replaces a difficult quantity by one that is easier to control while preserving the direction and the equality case that matter. The hard part is rarely remembering a named theorem. It is deciding which information may be discarded, which signs and domains make a transformation legal, and whether the resulting bound is sharp enough to finish the argument.

Why a bound can be more useful than an exact value #

In measurement and design, the exact disturbance is often unknowable while its maximum size is known. An inequality turns several uncertain contributions into a guarantee. The result may be less precise than an exact value, but it can still answer the practical question: is every permitted case safe or accurate enough?

Bounds are therefore decision tools. A worst-case bound can certify a design; an average or likely value cannot do that unless the decision itself is probabilistic.

Order-preserving transformations #

Start every proof with a domain and sign ledger. Addition preserves order without conditions, but multiplication, division, reciprocation, squaring, and applying a function do not.

An equivalence proof must justify both directions. Squaring $A\le B$ when the signs are unknown may introduce false solutions; multiplying two true inequalities is unsafe unless the signs of all factors are controlled.

Absolute value is distance, so for $r\ge0$,

$$|x-a|\le r \quad\Longleftrightarrow\quad a-r\le x\le a+r.$$

The triangle inequality and its reverse form are the basic tools for separating and comparing errors:

$$|x+y|\le |x|+|y|, \qquad \bigl||x|-|y|\bigr|\le |x-y|.$$

The same geometry survives in every norm: $\|u+v\|\le\|u\|+\|v\|$.

Cauchy–Schwarz turns correlation into energy #

For real or complex vectors $a,b$,

$$|\langle a,b\rangle|^2\le \langle a,a\rangle\langle b,b\rangle,$$

with equality exactly when the vectors are linearly dependent. In coordinates,

$$\left|\sum_{k=1}^{n}a_k\overline{b_k}\right|^2 \le \left(\sum_{k=1}^{n}|a_k|^2\right) \left(\sum_{k=1}^{n}|b_k|^2\right).$$

Why it works: remove from $a$ its projection onto $b$. The squared norm of the remainder cannot be negative. With $b\ne0$,

$$0\le\left\|a-\frac{\langle a,b\rangle}{\|b\|^2}b\right\|^2 =\|a\|^2-\frac{|\langle a,b\rangle|^2}{\|b\|^2}.$$

This proof explains both the inequality and equality: the remainder vanishes precisely when $a$ lies on the line spanned by $b$.

A frequently useful algebraic form follows by choosing $a_k=x_k/\sqrt{w_k}$ and $b_k=\sqrt{w_k}$, where $w_k>0$:

$$\sum_{k=1}^{n}\frac{x_k^2}{w_k} \ge \frac{(x_1+\cdots+x_n)^2}{w_1+\cdots+w_n}.$$

Means measure different kinds of size #

For positive $x_1,\ldots,x_n$,

$$H\le G\le A\le Q,$$

where

$$H=\frac{n}{\sum 1/x_k},\quad G=\left(\prod x_k\right)^{1/n},\quad A=\frac1n\sum x_k,\quad Q=\left(\frac1n\sum x_k^2\right)^{1/2}.$$

Equality throughout holds exactly when all $x_k$ are equal. That equality condition is often the design target: if the problem is symmetric and the expected extremum is $x_1=\cdots=x_n$, a means inequality is likely to preserve the right geometry.

Weighted AM–GM states that for $x_k>0$, $\lambda_k\ge0$, and $\sum\lambda_k=1$,

$$\prod_{k=1}^{n}x_k^{\lambda_k}\le\sum_{k=1}^{n}\lambda_kx_k.$$

The weights are not decoration. They let the exponents on the left match the monomial that must be bounded.

Hölder and the hierarchy of norms #

If $p,q>1$ and $1/p+1/q=1$, Hölder's inequality is

$$\sum_{k=1}^{n}|a_kb_k| \le \left(\sum |a_k|^p\right)^{1/p} \left(\sum |b_k|^q\right)^{1/q}.$$

Cauchy–Schwarz is the case $p=q=2$. Hölder is the right tool when factors naturally carry different powers; forcing everything into squares may lose the exponents that encode the problem.

Minkowski,

$$\left(\sum|a_k+b_k|^p\right)^{1/p} \le \left(\sum|a_k|^p\right)^{1/p} + \left(\sum|b_k|^p\right)^{1/p},$$

is the triangle inequality for the $\ell^p$ norm. Together, Hölder and Minkowski explain why many apparently unrelated inequalities are statements about dual norms and distance.

Convexity packages tangent and averaging bounds #

A function $\phi$ is convex on an interval when every chord lies above its graph:

$$\phi(\lambda x+(1-\lambda)y) \le \lambda\phi(x)+(1-\lambda)\phi(y),\qquad 0\le\lambda\le1.$$

If $\phi$ is differentiable, every tangent is a global lower bound on that interval:

$$\phi(y)\ge\phi(x)+\phi'(x)(y-x).$$

Jensen's inequality extends the chord property to finitely many weighted points:

$$\phi\!\left(\sum_{k=1}^{n}\lambda_kx_k\right) \le \sum_{k=1}^{n}\lambda_k\phi(x_k), \qquad \lambda_k\ge0,\qquad\sum\lambda_k=1.$$

For strictly convex $\phi$, equality holds only when all positively weighted $x_k$ are equal. For concave functions the direction reverses.

Choosing $\phi(x)=-\log x$, which is convex on $(0,\infty)$, yields weighted AM–GM after exponentiating. Choosing $\phi(x)=x^2$ gives the nonnegativity of variance:

$$\left(\sum\lambda_kx_k\right)^2\le\sum\lambda_kx_k^2.$$

Rearrangement and Chebyshev preserve order information #

If $a_1\le\cdots\le a_n$ and $b_1\le\cdots\le b_n$, the similarly ordered pairing maximizes $\sum a_kb_{\pi(k)}$ and the oppositely ordered pairing minimizes it. Chebyshev's sum inequality is a related average-correlation statement:

$$\frac1n\sum_{k=1}^{n}a_kb_k \ge \left(\frac1n\sum a_k\right) \left(\frac1n\sum b_k\right)$$

for similarly ordered real sequences; the direction reverses for oppositely ordered sequences. These tools are inappropriate when no ordering hypothesis has been established.

A proof-selection workflow #

1. Normalize the domain: write positivity, ordering, fixed sums/products, and forbidden denominators.
2. Rewrite the goal as a nonnegative gap, a norm bound, or a convexity statement.
3. Predict equality. A candidate theorem whose equality case cannot match is probably too loose.
4. Choose the invariant to preserve: sums suggest Cauchy–Schwarz; products suggest AM–GM; mixed powers suggest Hölder; averages through nonlinear functions suggest Jensen; ordered pairings suggest rearrangement.
5. Audit every direction change and every denominator.
6. Test boundary and equality cases. A numerical check cannot prove the result, but it quickly exposes a sign error or an impossible equality chain.