Cauchy-Schwarz Inequality

Cauchy Schwarz

The Cauchy Schwarz Inequality is

$$(\sqrt{a_1b_1}+\sqrt{a_2b_2}+\dots+\sqrt{a_nb_n})^2\le(a_1+a_2+\dots+a_n)(b_1+b_2+\dots+b_n)$$

for two variables. For three variables, it is

$$(\sqrt[3]{a_1b_1c_1}+\sqrt[3]{a_2b_2c_2}+\dots+\sqrt[3]{a_nb_nc_n})^3\le(a_1+a_2+\dots+a_n)(b_1+b_2+\dots+b_n)(c_1+c_2+\dots+c_n)$$

and similar for more variables.

It's useful because we can have very nifty cancellations. It helps separate variables that are connected, for example in

$$(\sqrt{ax}+\sqrt{by}+\sqrt{cz})^2\le(a+b+c)(x+y+z)$$

Here it separates $a$, $b$, and $c$ from $x$, $y$, and $z$.

Another application of it is in cancellations! For example,

$$(ax+by+cz)(\frac{a}x+\frac{b}y+\frac{c}z)\ge(\sqrt{ax\frac{a}x}+\sqrt{by\frac{b}{y}}+\sqrt{cz\frac{c}z})^2=(a+b+c)^2$$

There goes the $x$, $y$, and $z$!

Another property of it is that it's for all reals, not just nonnegative reals like AM-GM.

Note: a very helpful form of Cauchy Schwarz is in Titu's Lemma:

$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\dots+\frac{a_n^2}{b_n}\ge\frac{(a_1+a_2+\dots+a_n)^2}{b_1+b_2+\dots+b_n}$$

which follows from

$$\left(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\dots+\frac{a_n^2}{b_n}\right)(b_1+b_2+\dots+b_n)\ge\left(\sqrt{\frac{a_1^2}{b_1}b_1}+\sqrt{\frac{a_2^2}{b_2}b_2}+\dots+\sqrt{\frac{a_n^2}{b_n}b_n}\right)^2=(a_1+a_2+\dots+a_n)^2$$

This is a very helpful form because it can help combine denominators of fractions! My teacher always scolded me (rightfully) for writing $\frac{a}b+\frac{c}d=\frac{a+c}{b+d}$, but Titu's Lemma allows me to write $\frac{a}b+\frac{c}d\ge\frac{\sqrt{a}+\sqrt{c}}{b+d}$. TAKE THAT, TEACHER!

Inequalities (Part 4)

We're look at three useful inequalities in this part.

Cauchy Schwarz Inequality

For all real numbers $x_1, x_2, ..., x_n$ and $y_1, y_2, ..., y_n$ we have $(x_1^2 + x_2^2 + ... + x_n^2)(y_1^2 + y_2^2 + ... + y_n^2) \ge (x_1y_1 + x_2y_2 + ... + x_ny_n)^2$, with equality if and only if either $x_i = 0$ for all $i$, or there exists a number $i$ such that $\frac{y_i}{x_i} = t$ for all $i = 1, 2, ..., n$.

This is probably a very well-known inequality and can be proved by Lagrange's Identity easily. This inequality also implies the next, less famous, but nevertheless more useful form (for me) of Cauchy Schwarz.

Titu's Lemma

For all real numbers $x_1, x_2, ..., x_n$ and $y_1, y_2, ..., y_n$ we have

$\frac{x_1^2}{y_1} + \frac{x_2^2}{y_2} + ... + \frac{x_n^2}{y_n} \ge \frac{(x_1 + x_2 + ... + x_n)^2}{y_1 + y_2 + ... + y_n}$

One note that this is just Cauchy Schwarz after multiplying the denominator of the RHS to both sides.

Finally, the next inequality like a generalization of Cauchy Schwarz.

Holder's Inequality

Let $a_{11}, ..., a_{1n}, a_{21}, ..., a_{kn}$ be nonnegative real numbers. Then

$(a_{11} + a_{12} + ... + a_{1n})(a_{21} + a_{22} + ... + a_{2n})...(a_{k1} + a_{k2} + ... + a_{kn}) \ge (\sqrt[k]{a_{11}a_{21}....a_{k1}} + ... + \sqrt[k]{a_{1n}a_{2n}....a_{kn}})^k$.

This generalization of Cauchy Schwarz is really useful. The next part will have examples of these inequalities in action.

Inequalities (Part 5)

This part will consists of applications of Cauchy Schwarz, Holder's and Titu Lemma on some problems.

Prove that for all positive real numbers $a, b, c$,

$(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \ge (a + b + c)^3$.

Direct Holder unfortunately fails. However, we can use a trick to use Holder here. Note that $x^5 - x^2 + 3 \ge x^3 + 2$ is equivalent to $(x - 1)^2(x + 1)(x^2 + x + 1) \ge 0$, which is obviously true. Thus, we can now use Holder to find that $(a^3 + 1 + 1)(b^3 + 1 + 1)(c^3 + 1 + 1) \ge (a + b + c)^3$. This proves the inequality.

We continue with another problem, this time it's a proposal of me to Proofathon for the inequalities contest (that's over).

Suppose $a, b, c$ are positive reals such that $ab(a + b) + bc(b + c) + ca(c + a) + 2abc = 1$. Prove that $a^3(a + b)^2 + b^3(b + c)^2 + c^3(c + a)^2 \ge 6(a^2b + b^2c + c^2a)^3$.

Note that the equality is equivalent to $(a + b)(b + c)(c + a) = 1$.

Note that by Holder's Inequality, we have

$(a^5 + b^3c^2 + 2ac^4)(a + b)^2(b + c)^2 = (a^5 + b^3c^2 + 2ac^4)(2ab + b^2 + a^2)(b^2 + 2bc + c^2) \ge 2(a^2b + b^2c + c^2a)^3$.

Multiplying similar inequalities gives $\displaystyle\prod_{cyc}(a^5 + b^3c^2 + 2ac^4) \ge 8(a^2b + b^2c + c^2a)^9$, since $(a + b)(b + c)(c + a) = 1$.

Thus, $\displaystyle\sum_{cyc}(a^3(a + b)^2) = \displaystyle\sum_{cyc}(a^5 + b^3c^2 + 2ac^4) \ge 3\sqrt[3]{\displaystyle\prod_{cyc}(a^5 + b^3c^2 + 2ac^4)} \ge 6(a^2 b + b^2c + c^2a)^3$. (where the first inequality follows by AM-GM while the second follows by what we did above.

This proves the inequality.

Regrouping the terms then applying Holder is also a common technique.

You might be tired of Holder's Inequality already. Let's get on with one sweet application of Titu's Lemma.

Let $a, b, c$ be positive real numbers with $ab + bc + ca = \frac{1}{3}$. Show that

$\frac{a}{a^2 - bc + 1} + \frac{b}{b^2 - ca + 1} + \frac{c}{c^2 - ab + 1} \ge \frac{1}{a + b + c}$.

By Titu's Lemma,

$\frac{a}{a^2 - bc + 1} + \frac{b}{b^2 - ca + 1} + \frac{c}{c^2 - ab + 1}$ $= \frac{a^2}{a^3 - abc + a} + \frac{b^2}{b^3 - abc + b} + \frac{c^2}{c^3 - abc + c} \ge \frac{(a + b + c)^2}{(a^3 + b^3 + c^3) + (a + b + c) - 3abc}$ $= \frac{(a + b + c)^2}{(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca + 1)}$ $= \frac{a + b + c}{a^2 + b^2 + c^2 - ab - bc - ca + 3(ab + bc + ca)}$ $= \frac{1}{a + b + c}$, as desired.

An important trick to remember is to multiply both the numerator and the denominator by some term so that Titu's Lemma can be applied.

Finally, we end with a classical problem on Cauchy Schwarz.

(Iran 1998) Prove that if $x, y, z$ are positive reals such that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2$, then $\sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1} \le \sqrt{x + y + z}$.

We have by Cauchy Schwarz,

$\sqrt{(x + y + z)(\frac{x - 1}{x} + \frac{y - 1}{y} + \frac{z - 1}{z})} \ge \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}$

Since $\frac{x - 1}{x} + \frac{y - 1}{y} + \frac{z - 1}{z} = 3 - \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$, we're done.

This concludes the part.

Let $a$ and $b$ be positive real numbers with $a+b=1$. Prove that

$$\dfrac{a^2}{a+1}+\dfrac{b^2}{b+1}\geq \dfrac{1}{3}$$

(Source: Hungary)