Useful Inequalities for Analysis

A Small Toolkit for Estimating Real Quantities

In the previous tutorial, Bernoulli’s Inequality gave a lower bound for a power using its constant and linear terms. Analysis also relies on inequalities that control products and sums in terms of simpler quantities, especially squares. Such estimates let us replace an expression that is difficult to handle by one with a clear bound.

We begin with a bound for a product of two real numbers, then prove the finite-dimensional Cauchy–Schwarz inequality. The latter controls a sum of products by the sizes of the two lists of numbers. These results do not require limits or calculus; they follow from nonnegativity of squares and the order properties of the real numbers.

A Product Bound from a Square

For any real numbers $x$ and $y$, the square $(|x|-|y|)^2$ is nonnegative. Expanding it gives an immediate relation between the product $|xy|$ and the squares $x^2$ and $y^2$.

Theorem (Product Bound by Squares). For all $x,y\in\mathbb{R}$, $$ 2|xy|\leq x^2+y^2. $$ Equivalently, $$ |xy|\leq\frac{x^2+y^2}{2}. $$

Proof. Since squares are nonnegative,

0\leq(|x|-|y|)^2 =|x|^2-2|x||y|+|y|^2 =x^2-2|xy|+y^2.

Here $|x|^2=x^2$, $|y|^2=y^2$, and $|x||y|=|xy|$, using the absolute value properties established earlier in this course. Rearranging the final inequality gives $2|xy|\leq x^2+y^2$. Dividing by the positive number $2$ gives the equivalent form. $\square$

The absolute value is important. The product bound implies both $-\frac{x^2+y^2}{2}\leq xy$ and $xy\leq\frac{x^2+y^2}{2}$, because $-|xy|\leq xy\leq|xy|$. In applications, the upper bound alone may be enough, but using $|xy|$ gives control regardless of the signs of $x$ and $y$.

Worked Example: Bounding a Mixed Term

Use the product bound to estimate $6|uv|$ in terms of $u^2$ and $v^2$. Multiplying $2|uv|\leq u^2+v^2$ by the positive number $3$ preserves the inequality:

6|uv|\leq3u^2+3v^2.

For example, if $u=2$ and $v=-1$, then the left side is $6|2(-1)|=12$, while the right side is $3(2^2)+3((-1)^2)=12+3=15$. Thus the estimate gives $12\leq15$, as claimed. The bound need not be an equality for every choice of $u,v$; it is an upper estimate that works uniformly.

A Product Estimate with a Parameter

Sometimes the two square terms in a bound need different coefficients. A freely chosen positive parameter lets us adjust the estimate to suit the rest of a calculation. This flexibility is often more useful than the symmetric bound above.

Theorem (Parameterized Product Bound). For all $x,y\in\mathbb{R}$ and every $\varepsilon>0$, $$ |xy|\leq\varepsilon x^2+\frac{y^2}{4\varepsilon}. $$

Proof. The square of the real number $2\varepsilon|x|-|y|$ is nonnegative. Expanding gives

0\leq(2\varepsilon|x|-|y|)^2 =4\varepsilon^2x^2-4\varepsilon|xy|+y^2.

Because $\varepsilon>0$, division by $4\varepsilon$ preserves the inequality. The result is

0\leq\varepsilon x^2-|xy|+\frac{y^2}{4\varepsilon},

which rearranges to the stated bound. $\square$

The value of $\varepsilon$ is chosen by the needs of the estimate: making the coefficient of $x^2$ smaller makes the coefficient of $y^2$ larger. The inequality is valid for every positive choice, but not every choice will be equally useful. At $\varepsilon=\frac12$, it reduces to the symmetric product bound.

Worked Example: Choosing the Coefficient

Bound $5|uv|$ by an expression whose $u^2$ coefficient is $\frac12$. Apply the parameterized product bound to $|uv|$, choosing $\varepsilon=\frac{1}{10}$:

|uv|\leq\frac{1}{10}u^2+ \frac{v^2}{4(1/10)} =\frac{1}{10}u^2+\frac{5}{2}v^2.

Multiplying by $5>0$ gives

5|uv|\leq\frac12u^2+\frac{25}{2}v^2.

The coefficients follow directly from the choice of parameter: $5\varepsilon=\frac12$, and $5/(4\varepsilon)=5/(4/10)=\frac{25}{2}$. For $u=2$ and $v=1$, this reads $10\leq2+\frac{25}{2}=\frac{29}{2}$, which is true. The estimate is not intended to make both coefficients small at once; its purpose is to set one coefficient as needed while accepting the corresponding cost in the other.

The Cauchy–Schwarz Inequality

The product bound handles one pair of real numbers. For finite lists, the corresponding central estimate controls a sum of products by the sums of squares. This is the Cauchy–Schwarz inequality. It is a new tool for bounding a whole expression at once, rather than estimating its terms separately.

Theorem (Cauchy–Schwarz Inequality for Finite Sums). Let $n\geq1$ be an integer, and let $x_1,\ldots,x_n,y_1,\ldots,y_n\in\mathbb{R}$. Then $$ \left|\sum_{k=1}^{n}x_ky_k\right| \leq \left(\sum_{k=1}^{n}x_k^2\right)^{1/2} \left(\sum_{k=1}^{n}y_k^2\right)^{1/2}. $$

Proof. Set

A=\sum_{k=1}^{n}x_k^2,\qquad B=\sum_{k=1}^{n}x_ky_k,\qquad C=\sum_{k=1}^{n}y_k^2.

Each square in the definitions of $A$ and $C$ is nonnegative, so $A\geq0$ and $C\geq0$. If $C=0$, every $y_k^2$ must equal $0$: a finite sum of nonnegative terms can be zero only when every term is zero. Hence each $y_k=0$, so $B=0$, and the asserted inequality holds with both sides equal to $0$.

Now suppose $C>0$. For every real number $t$, each square $(x_k-ty_k)^2$ is nonnegative, and therefore

0\leq\sum_{k=1}^{n}(x_k-ty_k)^2 =A-2tB+t^2C.

Choose $t=B/C$, which is defined because $C>0$. Substitution gives

0\leq A-\frac{2B^2}{C}+\frac{B^2}{C} =A-\frac{B^2}{C}.

Multiplying by $C>0$ yields $B^2\leq AC$. Since both sides are nonnegative, comparison of absolute values by squares gives $|B|\leq\sqrt{AC}$. Finally, $\sqrt{AC}=\sqrt{A}\sqrt{C}$, since $A,C\geq0$. Substituting the definitions of $A,B,C$ proves the theorem. $\square$

The proof works by choosing $t$ to make the sum of squares as informative as possible. The case $C=0$ must be separated because $B/C$ would otherwise be undefined. This is a useful proof habit: before dividing by an expression, check whether it could be zero.

Worked Example: Bounding a Linear Combination

For real numbers $x,y,z$, estimate $|2x-y+2z|$ in terms of $x^2+y^2+z^2$. Apply Cauchy–Schwarz to the lists $(x,y,z)$ and $(2,-1,2)$:

|2x-y+2z| \leq\sqrt{x^2+y^2+z^2}\sqrt{2^2+(-1)^2+2^2} =3\sqrt{x^2+y^2+z^2}.

The coefficient-list calculation is $2^2+(-1)^2+2^2=4+1+4=9$, whose positive square root is $3$. As a numerical check, for $x=1,y=2,z=-1$, the left side is $|2-2-2|=2$, while the right side is $3\sqrt{1+4+1}=3\sqrt6$, which is greater than $2$. The bound applies to every real triple, not just this test.

Worked Example: Bounding an Average

Let $a,b,c\in\mathbb{R}$. Cauchy–Schwarz applied to the lists $(a,b,c)$ and $(1,1,1)$ gives

|a+b+c| \leq\sqrt{a^2+b^2+c^2}\sqrt{1^2+1^2+1^2} =\sqrt{3}\sqrt{a^2+b^2+c^2}.

Dividing by $3>0$ and using $\sqrt3/3=1/\sqrt3$, we obtain

\left|\frac{a+b+c}{3}\right| \leq\sqrt{\frac{a^2+b^2+c^2}{3}}.

For $a=2,b=-1,c=2$, the average is $1$, while the right side is $\sqrt{(4+1+4)/3}=\sqrt3$. Thus $1\leq\sqrt3$, in agreement with the estimate. The bound says that the absolute value of the average is controlled by the square root of the average of the squares.

Using the Estimates Carefully

These inequalities are useful because they replace a mixed expression by quantities whose signs and sizes are easier to control. For two terms, the product bound gives $xy\leq |xy|\leq(x^2+y^2)/2$. For a sum of many products, Cauchy–Schwarz can be more efficient than applying the triangle inequality and estimating every term separately. The triangle inequality remains available, but Cauchy–Schwarz often captures the combined structure of a sum.

An estimate can be correct and still be too weak for a particular purpose. The parameterized product bound illustrates this: decreasing the coefficient on $x^2$ increases the coefficient on $y^2$. Choose the parameter only after identifying which terms the surrounding argument can absorb. Likewise, check that every factor used to multiply or divide an inequality has the required sign.

The common foundation is nonnegativity of squares. The product estimate comes from one square, and the Cauchy–Schwarz proof comes from a sum of squares. In later arguments, these techniques let us control error terms, sums, and differences without computing them exactly. The key is to keep the hypotheses visible and to verify that the resulting bound is strong enough for the task at hand.

Key takeaway. For real $x,y$, $2|xy|\leq x^2+y^2$, and for every $\varepsilon>0$, $|xy|\leq\varepsilon x^2+y^2/(4\varepsilon)$. For finite lists, Cauchy–Schwarz bounds the absolute value of a sum of products by the product of the two square-root sums of squares.

Check Your Understanding

Use the square-based estimates and Cauchy–Schwarz to answer these questions.

Use the product bound to show that $4|xy|\leq2x^2+2y^2$ for all real $x,y$.
In the parameterized product bound, what value of $\varepsilon$ gives a coefficient of $1$ on $x^2$ when bounding $|xy|$? What is then the coefficient on $y^2$?
Apply Cauchy–Schwarz to bound $|x+3y|$ in terms of $x^2+y^2$.
Why does the proof of Cauchy–Schwarz treat the case $\sum_{k=1}^n y_k^2=0$ separately?
For which values of $x$ and $y$ does $2|xy|=x^2+y^2$?