Proving Inequalities: Problems with Solutions

Solution:
Complete the squares:
[tex]x^2+y^2+2x-4y+6=\left(x^2+2x+1\right)+\left(y^2-4y+4\right)+1=(x+1)^2+(y-2)^2+1[/tex]
Squares are non-negative, so the expression is at least [tex]1[/tex] for all real [tex]x[/tex] and [tex]y[/tex]:
[tex](x+1)^2+(y-2)^2+1\ge 1[/tex]
The value [tex]1[/tex] is attained for [tex]x=-1[/tex], [tex]y=2[/tex]. Hence the smallest value is [tex]1[/tex].

Solution:
For every [tex]k=1,2,\ldots,n[/tex] we have [tex]\sqrt{k}\le\sqrt{n}[/tex], hence
[tex]\frac{1}{\sqrt{k}}\ge\frac{1}{\sqrt{n}}[/tex]
Adding these [tex]n[/tex] inequalities, we get
[tex]\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}\ge n\times \frac{1}{\sqrt{n}}=\sqrt{n}[/tex]
which proves the inequality. Equality holds only for [tex]n=1[/tex]; for [tex]n\ge 2[/tex] the first term satisfies [tex]1\gt\frac{1}{\sqrt{n}}[/tex], so the inequality is strict.

Solution:
Move all terms to the left side. Writing [tex]-a^2=-2a^2+a^2[/tex] and [tex]2=1+1[/tex], we group the terms:
[tex]a^4+2-a^2-2a=\left(a^4-2a^2+1\right)+\left(a^2-2a+1\right)=\left(a^2-1\right)^2+\left(a-1\right)^2[/tex]
A sum of two squares is non-negative, so for every real [tex]a[/tex]
[tex]\left(a^2-1\right)^2+\left(a-1\right)^2\ge 0\Rightarrow a^4+2\ge a^2+2a[/tex]
Equality requires [tex]a^2-1=0[/tex] and [tex]a-1=0[/tex] at the same time, i.e. it holds only for [tex]a=1[/tex].

Solution:
Expand and use [tex]xy=2[/tex]:
[tex]\left(x+1\right)\left(y+2\right)=xy+2x+y+2=2x+y+4[/tex]
By the AM-GM inequality for the positive numbers [tex]2x[/tex] and [tex]y[/tex]:
[tex]2x+y\ge 2\sqrt{2x\times y}=2\sqrt{2\times 2}=4\Rightarrow \left(x+1\right)\left(y+2\right)\ge 4+4=8[/tex]
Equality requires [tex]2x=y[/tex]; together with [tex]xy=2[/tex] this gives [tex]2x^2=2[/tex], so [tex]x=1[/tex] (as [tex]x\gt 0[/tex]) and [tex]y=2[/tex]. Indeed, [tex]\left(1+1\right)\left(2+2\right)=8[/tex]. Hence the smallest value is [tex]8[/tex].

Solution:
The four factors are positive and their sum is constant:
[tex]\left(1+a\right)+\left(1+b\right)+\left(1+c\right)+\left(1+d\right)=4+\left(a+b+c+d\right)=5[/tex]
By the AM-GM inequality for these four positive numbers
[tex]\sqrt[4]{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\le\frac{5}{4}[/tex]
Raising both sides to the fourth power, we get
[tex]\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\le\left(\frac{5}{4}\right)^4=\frac{625}{256}[/tex]
Equality holds when all four factors are equal, i.e. for [tex]a=b=c=d=\frac{1}{4}[/tex], which satisfies the conditions. Hence the greatest value is [tex]\frac{625}{256}[/tex].

Solution:
Write the left side as a sum of five positive numbers:
[tex]2\sqrt{a}+3\sqrt[3]{b}=\sqrt{a}+\sqrt{a}+\sqrt[3]{b}+\sqrt[3]{b}+\sqrt[3]{b}[/tex]
By the AM-GM inequality for these five numbers
[tex]\sqrt{a}+\sqrt{a}+\sqrt[3]{b}+\sqrt[3]{b}+\sqrt[3]{b}\ge 5\sqrt[5]{\left(\sqrt{a}\right)^2\left(\sqrt[3]{b}\right)^3}=5\sqrt[5]{ab}[/tex]
which proves the inequality. Equality holds exactly when the five numbers are equal, i.e. [tex]\sqrt{a}=\sqrt[3]{b}[/tex], which means [tex]a^3=b^2[/tex] (for example [tex]a=b=1[/tex]).

Solution:
Put [tex]x=a^3[/tex], [tex]y=b^3[/tex], [tex]z=c^3[/tex]; these numbers are positive and [tex]xyz=\left(abc\right)^3=1[/tex]. Expand the left side:
[tex]\left(x+2\right)\left(y+2\right)\left(z+2\right)=xyz+2\left(xy+yz+zx\right)+4\left(x+y+z\right)+8=2\left(xy+yz+zx\right)+4\left(x+y+z\right)+9[/tex]
On the right side each product has the form [tex]\left(x+2\right)\left(y+2\right)=xy+2x+2y+4[/tex], and each variable occurs in two of the three products, so the right side equals
[tex]\left(xy+yz+zx\right)+4\left(x+y+z\right)+12[/tex]
The difference between the left and the right side is [tex]\left(xy+yz+zx\right)-3[/tex], so it suffices to prove that [tex]xy+yz+zx\ge 3[/tex]. By the AM-GM inequality for the positive numbers [tex]xy[/tex], [tex]yz[/tex], [tex]zx[/tex]:
[tex]xy+yz+zx\ge 3\sqrt[3]{xy\times yz\times zx}=3\sqrt[3]{\left(xyz\right)^2}=3[/tex]
This proves the inequality. Equality holds when [tex]xy=yz=zx[/tex], i.e. [tex]x=y=z=1[/tex], i.e. [tex]a=b=c=1[/tex].

Solution:
Consider the sum of the squares of the deviations from [tex]n[/tex] and expand it:
[tex]\left(a_1-n\right)^2+\ldots+\left(a_n-n\right)^2=\left(a_1^2+\ldots+a_n^2\right)-2n\left(a_1+\ldots+a_n\right)+n\times n^2[/tex]
By the second condition the first bracket is at most [tex]n^3+1[/tex]. By the first condition [tex]a_1+\ldots+a_n\ge n^2[/tex], and since [tex]-2n\lt 0[/tex], the middle term is at most [tex]-2n\times n^2=-2n^3[/tex]. Hence
[tex]\left(a_1-n\right)^2+\ldots+\left(a_n-n\right)^2\le n^3+1-2n^3+n^3=1[/tex]
All squares are non-negative, so each of them is at most the whole sum. Hence for every [tex]k[/tex]
[tex]\left(a_k-n\right)^2\le 1\iff\left|a_k-n\right|\le 1\iff -1\le a_k-n\le 1\iff n-1\le a_k\le n+1[/tex]
which is what we had to prove.