Practice
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.