Chapter
Theory
An inequality in olympiad algebra is rarely proved by visual guessing. Usually it is reduced to an obvious nonnegative quantity: a square, a sum of squares, AM-GM, Cauchy-Schwarz, or a careful rearrangement of terms.
The main habit in this module is to check the equality case, not just the sign. Equality often suggests the right substitution and shows how sharp the estimate is.
For all real \(a,b\), \((a-b)^2\ge0\), hence \(a^2+b^2\ge2ab\). For positive numbers \(x_1,\ldots,x_n\), AM-GM says \[\frac{x_1+\cdots+x_n}{n}\ge\sqrt[n]{x_1\cdots x_n}.\]
A basic form of Cauchy-Schwarz is \[(a^2+b^2)(c^2+d^2)\ge(ac+bd)^2.\] In Engel form, one often uses \[\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{(x+y)^2}{a+b}\] for \(a,b>0\).
AM-GM is natural when positive quantities have a fixed sum or product. Cauchy works well with sums of fractions like \(\frac{x^2}{a}\). Sums of squares appear when expanding leaves terms such as \((a-b)^2\).
If a problem asks for a maximum or minimum, first find the likely equality case, then choose an inequality whose equality occurs exactly there.
If the expression is symmetric in \(a,b,c\), try to reduce it to \((a-b)^2+(b-c)^2+(c-a)^2\). If there are positive fractions, check whether Cauchy in Engel form applies. If the product \(abc\) appears and the sum is fixed, AM-GM is usually nearby.
For rearrangement problems, sort two sequences in the same order. The difference between the “right” and “wrong” arrangement often becomes a product of two nonnegative differences.
AM-GM cannot be applied to negative numbers. In fractional inequalities, denominators must be positive. Multiplying an inequality by an expression of unknown sign may reverse the inequality and lose the solution.
Another common mistake is proving a weaker inequality and not noticing that it does not imply the target. Always check numerically whether the chosen estimate is strong enough.
1. Are all quantities positive? 2. What is the equality case? 3. Can the difference be written as a sum of squares? 4. Is there a form \(\sum \frac{x^2}{a}\)? 5. Should the variables be sorted?
Examples
The basic idea is to prove an inequality through obvious nonnegativity.
Problem. Prove that \(a^2+b^2\ge2ab\) for all real \(a,b\).
Since \((a-b)^2\ge0\), we have \(a^2-2ab+b^2\ge0\). Hence \(a^2+b^2\ge2ab\). Equality holds only when \(a=b\).
This example shows how to find a minimum of an expression with a reciprocal.
Problem. For \(x>0\), prove that \(x+\frac{1}{x}\ge2\).
By AM-GM, \(x+\frac{1}{x}\ge2\sqrt{x\cdot\frac{1}{x}}=2\). Equality holds at \(x=1\).
If the sum of positive numbers is fixed, the product is often bounded above.
Problem. Let \(a,b,c>0\) and \(a+b+c=6\). Prove that \(abc\le8\).
By AM-GM, \(\frac{a+b+c}{3}\ge\sqrt[3]{abc}\). Hence \(2\ge\sqrt[3]{abc}\), so \(abc\le8\). Equality holds at \(a=b=c=2\).
Engel form turns a sum of fractions into the square of a sum.
Problem. For \(x,y>0\), prove that \(\frac{x^2}{y}+\frac{y^2}{x}\ge x+y\).
By Cauchy, \(\frac{x^2}{y}+\frac{y^2}{x}\ge\frac{(x+y)^2}{x+y}=x+y\). Equality holds when \(x=y\).
Cauchy often proves a lower bound for reciprocals.
Problem. Let \(x,y,z>0\) and \(x+y+z=1\). Prove that \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge9\).
By Cauchy, \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+y+z)\ge(1+1+1)^2=9\). Since \(x+y+z=1\), the claim follows. Equality holds when \(x=y=z=\frac{1}{3}\).
This is a standard first example where Cauchy is stronger than direct addition.
Problem. For \(a,b,c>0\), prove that \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\).
Write \(\frac{a}{b+c}=\frac{a^2}{a(b+c)}\). By Cauchy, the sum is at least \(\frac{(a+b+c)^2}{a(b+c)+b(c+a)+c(a+b)}\). The denominator is \(2(ab+bc+ca)\), and \((a+b+c)^2\ge3(ab+bc+ca)\). Thus the result is \(\frac{3}{2}\).
When two sets are ordered in the same way, the sum of paired products is largest.
Problem. If \(a\le b\) and \(x\le y\), prove that \(ay+bx\le ax+by\).
The difference between the right and left sides is \(ax+by-ay-bx=(b-a)(y-x)\ge0\). Therefore \(ay+bx\le ax+by\).
A strong solution usually explains immediately when equality can occur.
Problem. Let \(x,y,z>0\) and \(xyz=1\). Prove that \((1+x)(1+y)(1+z)\ge8\).
By AM-GM, \(1+x\ge2\sqrt{x}\), \(1+y\ge2\sqrt{y}\), \(1+z\ge2\sqrt{z}\). Multiplying, \((1+x)(1+y)(1+z)\ge8\sqrt{xyz}=8\). Equality holds when \(x=y=z=1\).
Problems
Prove that for all real \(a,b\), \(a^2+b^2\ge2ab\).
Start with \((a-b)^2\ge0\).
We have \((a-b)^2=a^2-2ab+b^2\ge0\). Hence \(a^2+b^2\ge2ab\). Equality occurs when \(a=b\).
For \(x>0\), find the least value of \(x+\frac{1}{x}\).
Apply AM-GM to \(x\) and \(\frac{1}{x}\).
By AM-GM, \(x+\frac{1}{x}\ge2\sqrt{x\cdot\frac{1}{x}}=2\). Equality holds at \(x=1\), so the minimum is \(2\).
Prove that for all real \(a,b,c\), \(a^2+b^2+c^2\ge ab+bc+ca\).
Multiply the difference by \(2\) and group squares.
\(2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2\ge0\). Therefore the inequality holds. Equality occurs when \(a=b=c\).
Let \(x,y>0\) and \(x+y=10\). Prove that \(xy\le25\).
Apply AM-GM to \(x\) and \(y\).
\(\frac{x+y}{2}\ge\sqrt{xy}\), so \(5\ge\sqrt{xy}\). Hence \(xy\le25\). Equality occurs when \(x=y=5\).
Let \(a,b,c>0\) and \(a+b+c=6\). Prove that \(abc\le8\).
Use AM-GM for three positive numbers.
By AM-GM, \(\frac{a+b+c}{3}\ge\sqrt[3]{abc}\). The left side is \(2\), so \(\sqrt[3]{abc}\le2\), hence \(abc\le8\). Equality occurs when \(a=b=c=2\).
For \(a,b>0\), prove that \(\frac{a}{b}+\frac{b}{a}\ge2\).
Apply AM-GM to the two positive fractions.
By AM-GM, \(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}\cdot\frac{b}{a}}=2\). Equality holds when \(a=b\).
Let \(x,y>0\). Prove that \(\frac{x^2}{y}+\frac{y^2}{x}\ge x+y\).
Use Engel form: \(\sum \frac{u_i^2}{v_i}\ge\frac{(\sum u_i)^2}{\sum v_i}\).
By Cauchy, \(\frac{x^2}{y}+\frac{y^2}{x}\ge\frac{(x+y)^2}{x+y}=x+y\). Equality holds when \(x=y\).
For \(a,b>0\), prove that \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{(x+y)^2}{a+b}\) for all real \(x,y\).
This is exactly Cauchy in Engel form.
By Cauchy-Schwarz in Engel form, \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{(x+y)^2}{a+b}\). Equality occurs when \(\frac{x}{a}=\frac{y}{b}\).
Let \(a,b,c\ge0\) and \(a+b+c=1\). Prove that \(ab+bc+ca\le\frac{1}{3}\).
Use \((a+b+c)^2\ge3(ab+bc+ca)\).
From the three-squares inequality, \((a+b+c)^2\ge3(ab+bc+ca)\). Since \(a+b+c=1\), we get \(1\ge3(ab+bc+ca)\), so \(ab+bc+ca\le\frac{1}{3}\). Equality holds at \(a=b=c=\frac{1}{3}\).
Let \(x,y,z>0\) and \(xyz=1\). Prove that \(x+y+z\ge3\).
Apply AM-GM to \(x,y,z\).
By AM-GM, \(\frac{x+y+z}{3}\ge\sqrt[3]{xyz}=1\). Therefore \(x+y+z\ge3\). Equality occurs when \(x=y=z=1\).
Let \(a,b,c>0\). Prove \[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}.\]
Write \(\frac{a}{b+c}\) as \(\frac{a^2}{a(b+c)}\) and apply Cauchy.
By Cauchy, the sum is at least \(\frac{(a+b+c)^2}{a(b+c)+b(c+a)+c(a+b)}=\frac{(a+b+c)^2}{2(ab+bc+ca)}\).
Since \((a+b+c)^2\ge3(ab+bc+ca)\), we get \(\frac{(a+b+c)^2}{2(ab+bc+ca)}\ge\frac{3}{2}\). Equality holds when \(a=b=c\).
Let \(x,y,z>0\) and \(x+y+z=1\). Prove that \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge9\).
Apply Cauchy to \(\left(\sum \frac{1}{x}\right)(x+y+z)\).
By Cauchy, \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x+y+z)\ge(1+1+1)^2=9\). Since \(x+y+z=1\), the inequality follows. Equality holds when \(x=y=z=\frac{1}{3}\).
Let \(a,b,c>0\). Prove that \(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c\).
Apply Cauchy in Engel form to the three fractions.
By Cauchy, \(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{(a+b+c)^2}{a+b+c}=a+b+c\). Equality holds when \(a=b=c\).
Let \(x,y,z>0\) and \(xyz=1\). Prove that \((1+x)(1+y)(1+z)\ge8\).
Estimate each factor \(1+x\) by \(2\sqrt{x}\).
By AM-GM, \(1+x\ge2\sqrt{x}\), \(1+y\ge2\sqrt{y}\), \(1+z\ge2\sqrt{z}\). Multiplying, \((1+x)(1+y)(1+z)\ge8\sqrt{xyz}=8\). Equality holds when \(x=y=z=1\).
Let \(a,b,c\) be real numbers and \(a+b+c=0\). Prove that \(a^2+b^2+c^2\ge0\), with equality only when \(a=b=c=0\). Then explain why this implies \(x^2+y^2+z^2\ge\frac{(x+y+z)^2}{3}\).
In the second part, subtract the average from \(x,y,z\).
The first part is obvious because a sum of squares is nonnegative; equality is possible only when all squares are zero.
For the second part, set \(m=\frac{x+y+z}{3}\), \(a=x-m\), \(b=y-m\), \(c=z-m\). Then \(a+b+c=0\), so \((x-m)^2+(y-m)^2+(z-m)^2\ge0\). Expanding gives \(x^2+y^2+z^2\ge3m^2=\frac{(x+y+z)^2}{3}\).
Let \(a\le b\le c\) and \(x\le y\le z\). Prove that \(ax+by+cz\ge az+by+cx\).
Cancel the common term \(by\) and look at the difference.
The difference is \(ax+cz-az-cx=(c-a)(z-x)\). Since \(c-a\ge0\) and \(z-x\ge0\), it is nonnegative. Therefore \(ax+by+cz\ge az+by+cx\).
Let \(x,y,z>0\). Prove \[\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac{x+y+z}{2}.\]
Apply Cauchy in Engel form to the whole sum.
By Cauchy, the left-hand side is at least \(\frac{(x+y+z)^2}{(y+z)+(z+x)+(x+y)}=\frac{(x+y+z)^2}{2(x+y+z)}=\frac{x+y+z}{2}\). Equality holds when \(x=y=z\).
Let \(a,b,c>0\) and \(a+b+c=3\). Prove \[\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c}\ge\frac{3}{4}.\]
Use Cauchy: \(\sum \frac{1^2}{3+a}\ge\frac{9}{(3+a)+(3+b)+(3+c)}\).
By Cauchy, the left-hand side is at least \(\frac{(1+1+1)^2}{(3+a)+(3+b)+(3+c)}=\frac{9}{9+a+b+c}=\frac{9}{12}=\frac{3}{4}\). Equality holds when \(a=b=c=1\).
Let \(a,b,c>0\) and \(a+b+c=1\). Prove \[\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\ge\frac{3}{2}.\]
Replace \(1-a\) by \(b+c\), and similarly.
Since \(a+b+c=1\), we have \(1-a=b+c\), \(1-b=c+a\), \(1-c=a+b\). Thus the inequality becomes \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\), which is Nesbitt's inequality.
Let \(x,y,z>0\) and \(x+y+z=6\). Prove \[\frac{x^2}{x+1}+\frac{y^2}{y+1}+\frac{z^2}{z+1}\ge4.\]
Again apply Cauchy in Engel form and add the denominators.
By Cauchy, the left-hand side is at least \(\frac{(x+y+z)^2}{(x+1)+(y+1)+(z+1)}=\frac{36}{9}=4\). Equality holds when \(x=y=z=2\).
Let \(a,b,c>0\). Prove \[(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9.\]
This is a direct application of Cauchy-Schwarz.
By Cauchy, \((a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge(\sqrt{a}\cdot\frac{1}{\sqrt{a}}+\sqrt{b}\cdot\frac{1}{\sqrt{b}}+\sqrt{c}\cdot\frac{1}{\sqrt{c}})^2=9\). Equality holds when \(a=b=c\).
Let \(a,b,c>0\). Prove \[\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{a+b+c}{2}.\]
Apply Cauchy and carefully add the denominators.
By Cauchy, the left-hand side is at least \(\frac{(a+b+c)^2}{(a+b)+(b+c)+(c+a)}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}\).
Let \(a,b,c>0\) and \(abc=1\). Prove \[(a+b+1)(b+c+1)(c+a+1)\ge27.\]
Show that each factor is at least \(3\) using AM-GM.
By AM-GM, \(a+b+1\ge3\sqrt[3]{ab}\), \(b+c+1\ge3\sqrt[3]{bc}\), \(c+a+1\ge3\sqrt[3]{ca}\). Multiplying, the left-hand side is at least \(27\sqrt[3]{a^2b^2c^2}=27\), since \(abc=1\). Equality holds when \(a=b=c=1\).
Let \(a,b,c\ge0\). Prove \[a^3+b^3+c^3+3abc\ge a^2b+a^2c+b^2a+b^2c+c^2a+c^2b.\]
Assume without loss of generality that \(a\ge b\ge c\), and consider \(\sum a(a-b)(a-c)\).
The difference between the left and right sides is \(\sum a(a-b)(a-c)\). Assume \(a\ge b\ge c\). Then \[\sum a(a-b)(a-c)=(a-b)^2(a+b-c)+c(a-c)(b-c).\]
Both terms are nonnegative: \(a-b\ge0\), \(a+b-c\ge0\), \(c\ge0\), \(a-c\ge0\), \(b-c\ge0\). Hence the whole sum is nonnegative, and the inequality is proved.
Ladders