Proving Inequalities: Difficult Problems with Solutions

Solution:
The denominator [tex]x^4+y^4[/tex] is positive, so the inequality [tex]\frac{xy^3+x^3y}{x^4+y^4}\ge -1[/tex] is equivalent to [tex]x^4+y^4+x^3y+xy^3\ge 0[/tex]. Factor the left side:
[tex]x^4+x^3y+xy^3+y^4=x^3\left(x+y\right)+y^3\left(x+y\right)=\left(x+y\right)\left(x^3+y^3\right)=\left(x+y\right)^2\left(x^2-xy+y^2\right)[/tex]
Here [tex]\left(x+y\right)^2\ge 0[/tex], and since [tex]y\ne 0[/tex],
[tex]x^2-xy+y^2=\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}\gt 0[/tex]
So the product is non-negative and the expression is always at least [tex]-1[/tex]. The value [tex]-1[/tex] is attained when [tex]x+y=0[/tex], for example for [tex]x=1[/tex], [tex]y=-1[/tex]:
[tex]\frac{1\times \left(-1\right)^3+1^3\times \left(-1\right)}{1^4+\left(-1\right)^4}=\frac{-2}{2}=-1[/tex]
Hence the smallest value is [tex]-1[/tex].

Solution:
Put [tex]x=b+c[/tex], [tex]y=c+a[/tex], [tex]z=a+b[/tex]; these numbers are positive. Since [tex]a+b+c=1[/tex], every factor can be written through them:
[tex]1-a=b+c=x,\quad 1-b=y,\quad 1-c=z[/tex]
[tex]1+a=\left(a+b+c\right)+a=\left(a+b\right)+\left(a+c\right)=y+z,\quad 1+b=z+x,\quad 1+c=x+y[/tex]
So the inequality becomes
[tex]\left(y+z\right)\left(z+x\right)\left(x+y\right)\ge 8xyz[/tex]
By the AM-GM inequality
[tex]y+z\ge 2\sqrt{yz},\quad z+x\ge 2\sqrt{zx},\quad x+y\ge 2\sqrt{xy}[/tex]
Both sides of these inequalities are positive, so they can be multiplied:
[tex]\left(y+z\right)\left(z+x\right)\left(x+y\right)\ge 8\sqrt{x^2y^2z^2}=8xyz[/tex]
which is what we had to prove. Equality holds for [tex]x=y=z[/tex], i.e. for [tex]a=b=c=\frac{1}{3}[/tex].

Solution:
Using [tex]a+b+c=1[/tex], write the first factor as
[tex]1+\frac{1}{a}=\frac{a+1}{a}=\frac{a+a+b+c}{a}[/tex]
By the AM-GM inequality for the four positive numbers [tex]a[/tex], [tex]a[/tex], [tex]b[/tex], [tex]c[/tex]
[tex]a+a+b+c\ge 4\sqrt[4]{a^2bc}\Rightarrow 1+\frac{1}{a}\ge\frac{4\sqrt[4]{a^2bc}}{a}[/tex]
In the same way
[tex]1+\frac{1}{b}\ge\frac{4\sqrt[4]{ab^2c}}{b},\quad 1+\frac{1}{c}\ge\frac{4\sqrt[4]{abc^2}}{c}[/tex]
Multiplying these three inequalities between positive numbers, we get
[tex]\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\ge\frac{64\sqrt[4]{a^4b^4c^4}}{abc}=64[/tex]
For [tex]a=b=c=\frac{1}{3}[/tex] the expression equals [tex]4\times 4\times 4=64[/tex]. Hence the smallest value is [tex]64[/tex].

Solution:
Denote
[tex]P=\frac{1}{2}\times \frac{3}{4}\times \ldots\times \frac{99}{100},\qquad Q=\frac{2}{3}\times \frac{4}{5}\times \ldots\times \frac{100}{101}[/tex]
Each product has [tex]50[/tex] positive factors. For every natural [tex]k[/tex] we have [tex]\frac{2k-1}{2k}\lt\frac{2k}{2k+1}[/tex], because after multiplying by the positive denominators this becomes
[tex]\left(2k-1\right)\left(2k+1\right)\lt 4k^2\iff 4k^2-1\lt 4k^2[/tex]
Multiplying these [tex]50[/tex] inequalities between positive numbers (for [tex]k=1,2,\ldots,50[/tex]), we get [tex]P\lt Q[/tex]. On the other hand,
[tex]P\times Q=\frac{1\times 2\times 3\times \ldots\times 100}{2\times 3\times 4\times \ldots\times 101}=\frac{1}{101}[/tex]
Hence
[tex]P^2\lt P\times Q=\frac{1}{101}\lt\frac{1}{100}[/tex]
and since [tex]P\gt 0[/tex], it follows that [tex]P\lt\frac{1}{10}[/tex], which is what we had to prove.

Solution:
Let [tex]t=xy[/tex]. From the condition, [tex]x^2+y^2=1-t[/tex], so
[tex]x^3y+xy^3=xy\left(x^2+y^2\right)=t\left(1-t\right)[/tex]
Now bound [tex]t[/tex]. From [tex]x^2+y^2\ge 2xy[/tex] and [tex]x^2+y^2\ge -2xy[/tex] we get
[tex]1-t\ge 2t\Rightarrow t\le\frac{1}{3},\qquad 1-t\ge -2t\Rightarrow t\ge -1[/tex]
So [tex]t\in\left[-1, \frac{1}{3}\right][/tex]. The graph of [tex]g\left(t\right)=t-t^2[/tex] is a parabola with vertex at [tex]t=\frac{1}{2}[/tex], and the function is increasing for [tex]t\le\frac{1}{2}[/tex]. The vertex is not admissible, because [tex]t\le\frac{1}{3}[/tex], so the value [tex]\frac{1}{4}[/tex] cannot be reached. On [tex]\left[-1, \frac{1}{3}\right][/tex] the maximum is
[tex]g\left(\frac{1}{3}\right)=\frac{1}{3}\times \frac{2}{3}=\frac{2}{9}[/tex]
It is attained for [tex]x=y=\frac{1}{\sqrt{3}}[/tex]: then [tex]x^2+xy+y^2=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1[/tex] and [tex]x^3y+xy^3=2\times \frac{1}{9}=\frac{2}{9}[/tex]. Hence the greatest value is [tex]\frac{2}{9}[/tex].

Solution:
For an acute angle [tex]s=\sin \alpha\gt 0[/tex] and [tex]c=\cos\alpha\gt 0[/tex]. Expanding the brackets:
[tex]\left(1+\frac{1}{s}\right)\left(1+\frac{1}{c}\right)=1+\frac{1}{s}+\frac{1}{c}+\frac{1}{sc}[/tex]
By the AM-GM inequality [tex]\frac{1}{s}+\frac{1}{c}\ge\frac{2}{\sqrt{sc}}[/tex], so
[tex]1+\frac{1}{s}+\frac{1}{c}+\frac{1}{sc}\ge 1+\frac{2}{\sqrt{sc}}+\frac{1}{sc}=\left(1+\frac{1}{\sqrt{sc}}\right)^2[/tex]
Moreover
[tex]sc=\sin \alpha\cos\alpha=\frac{\sin 2\alpha}{2}\le\frac{1}{2}\Rightarrow\frac{1}{\sqrt{sc}}\ge\sqrt{2}[/tex]
hence the expression is at least
[tex]\left(1+\sqrt{2}\right)^2=3+2\sqrt{2}[/tex]
Both estimates become equalities for [tex]\alpha=45^\circ[/tex], where [tex]s=c=\frac{\sqrt{2}}{2}[/tex]. Hence the smallest value is [tex]3+2\sqrt{2}[/tex].
Note that the bound [tex]4\sqrt{2}[/tex], obtained from [tex]1+\frac{1}{s}\ge\frac{2}{\sqrt{s}}[/tex] and [tex]1+\frac{1}{c}\ge\frac{2}{\sqrt{c}}[/tex], is never attained, because it would require [tex]\sin \alpha=\cos\alpha=1[/tex].

Solution:
Consider the vectors [tex]\vec{u}=\left(x, 1\right)[/tex], [tex]\vec{v}=\left(y, 2\right)[/tex] and [tex]\vec{w}=\left(z, 3\right)[/tex]. Their lengths are [tex]\sqrt{x^2+1}[/tex], [tex]\sqrt{y^2+4}[/tex] and [tex]\sqrt{z^2+9}[/tex], so the expression equals [tex]\left|\vec{u}\right|+\left|\vec{v}\right|+\left|\vec{w}\right|[/tex]. By the triangle inequality for vectors
[tex]\left|\vec{u}\right|+\left|\vec{v}\right|+\left|\vec{w}\right|\ge\left|\vec{u}+\vec{v}+\vec{w}\right|[/tex]
where [tex]\vec{u}+\vec{v}+\vec{w}=\left(x+y+z, 1+2+3\right)=\left(8, 6\right)[/tex], and the length of this vector is
[tex]\sqrt{8^2+6^2}=\sqrt{100}=10[/tex]
Hence the expression is always at least [tex]10[/tex]. Equality holds when the three vectors have the same direction, i.e. when their coordinates are proportional:
[tex]\frac{x}{1}=\frac{y}{2}=\frac{z}{3}=\lambda[/tex]
Then [tex]x+y+z=6\lambda=8[/tex], so [tex]\lambda=\frac{4}{3}[/tex] and [tex]x=\frac{4}{3}[/tex], [tex]y=\frac{8}{3}[/tex], [tex]z=4[/tex]. Check:
[tex]\sqrt{\frac{16}{9}+1}+\sqrt{\frac{64}{9}+4}+\sqrt{16+9}=\frac{5}{3}+\frac{10}{3}+5=10[/tex]
So the smallest value is [tex]10[/tex].

Solution:
Put [tex]u=tx[/tex], [tex]v=ty[/tex], [tex]w=tz[/tex]. Since [tex]\left|t\right|\le 1[/tex] and [tex]\left|x\right|, \left|y\right|, \left|z\right|\le 1[/tex], we have [tex]\left|u\right|, \left|v\right|, \left|w\right|\le 1[/tex], hence
[tex]1+u\ge 0,\quad 1+v\ge 0,\quad 1+w\ge 0[/tex]
In the new variables the left side is
[tex]t^2\left(xy+yz+zx\right)+2t\left(x+y+z\right)+3=uv+vw+wu+2\left(u+v+w\right)+3[/tex]
On the other hand, [tex]\left(1+u\right)\left(1+v\right)=1+u+v+uv[/tex], and adding the three analogous products gives
[tex]\left(1+u\right)\left(1+v\right)+\left(1+v\right)\left(1+w\right)+\left(1+w\right)\left(1+u\right)=3+2\left(u+v+w\right)+uv+vw+wu[/tex]
Therefore the left side of the inequality equals this sum of three products of non-negative numbers, so it is non-negative, which proves the inequality. Equality is possible, e.g. for [tex]t=1[/tex], [tex]x=y=z=-1[/tex]: [tex]3-6+3=0[/tex].

Solution:
Denote the expression by [tex]E[/tex]. Lower bound:
[tex]3\left(a^2+b^2+c^2\right)-\left(a+b+c\right)^2=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge 0[/tex]
so [tex]E\ge\frac{1}{3}[/tex], with equality for [tex]a=b=c[/tex] (equilateral triangle); thus the value [tex]\frac{1}{3}[/tex] is attained.
Upper bound: by the triangle inequality [tex]a\lt b+c[/tex]; multiplying by [tex]a\gt 0[/tex] gives [tex]a^2\lt ab+ac[/tex], and similarly [tex]b^2\lt ba+bc[/tex], [tex]c^2\lt ca+cb[/tex]. Adding these,
[tex]a^2+b^2+c^2\lt 2\left(ab+bc+ca\right)[/tex]
hence
[tex]\left(a+b+c\right)^2=\left(a^2+b^2+c^2\right)+2\left(ab+bc+ca\right)\gt 2\left(a^2+b^2+c^2\right)\Rightarrow E\lt\frac{1}{2}[/tex]
so the value [tex]\frac{1}{2}[/tex] is never attained. Every value [tex]v\in\left[\frac{1}{3}, \frac{1}{2}\right)[/tex] is attained. Take the isosceles triangle with sides [tex]1[/tex], [tex]1[/tex], [tex]t[/tex], where [tex]0\lt t\le 1[/tex] (it exists because [tex]t\lt 2[/tex]); then [tex]E=\frac{t^2+2}{\left(t+2\right)^2}[/tex], and the equation [tex]E=v[/tex] is equivalent to
[tex]\left(1-v\right)t^2-4vt+2-4v=0,\qquad \frac{D}{4}=4v^2-\left(1-v\right)\left(2-4v\right)=2\left(3v-1\right)\ge 0[/tex]
Its smaller root
[tex]t=\frac{2v-\sqrt{2\left(3v-1\right)}}{1-v}[/tex]
is positive, because [tex]4v^2-2\left(3v-1\right)=2\left(2v-1\right)\left(v-1\right)\gt 0[/tex] for [tex]v\lt\frac{1}{2}[/tex]; and [tex]t\le 1[/tex] is equivalent to [tex]3v-1\le\sqrt{2\left(3v-1\right)}[/tex], which holds since [tex]0\le 3v-1\lt 2[/tex]. Therefore the set of values is [tex]\left[\frac{1}{3}, \frac{1}{2}\right)[/tex].

Solution:
Denote [tex]S=a_1+a_2+\ldots+a_6[/tex] and [tex]P=a_1a_6+a_2a_5+a_3a_4[/tex], and consider the quadratic function
[tex]f\left(x\right)=\left(x-a_1\right)\left(x-a_6\right)+\left(x-a_2\right)\left(x-a_5\right)+\left(x-a_3\right)\left(x-a_4\right)=3x^2-Sx+P[/tex]
Its discriminant is [tex]D=S^2-12P[/tex], so we must prove that [tex]D\gt 0[/tex]. Evaluate [tex]f[/tex] at [tex]x=a_3[/tex]:
[tex]f\left(a_3\right)=\left(a_3-a_1\right)\left(a_3-a_6\right)+\left(a_3-a_2\right)\left(a_3-a_5\right)+0[/tex]
Since [tex]a_3-a_1\gt 0[/tex] and [tex]a_3-a_6\lt 0[/tex], the first product is negative; since [tex]a_3-a_2\gt 0[/tex] and [tex]a_3-a_5\lt 0[/tex], the second product is negative as well. Hence [tex]f\left(a_3\right)\lt 0[/tex]. But
[tex]f\left(x\right)=3\left(x-\frac{S}{6}\right)^2-\frac{D}{12}[/tex]
so if [tex]D\le 0[/tex], then [tex]f\left(x\right)\ge 0[/tex] for every [tex]x[/tex], a contradiction. Therefore [tex]D=S^2-12P\gt 0[/tex], which is the required inequality.

Solution:
Let [tex]S=a+b+c[/tex]. If [tex]a=b=c=0[/tex], the left side is [tex]0\lt 2[/tex], so [tex]S\gt 0[/tex]. Since [tex]b, c\in\left[0, 1\right][/tex], we have
[tex]\left(1-b\right)\left(1-c\right)\ge 0\Rightarrow b+c\le 1+bc[/tex]
Also [tex]a\le 1\le 1+bc[/tex]. Adding, [tex]S\le 2\left(1+bc\right)[/tex], hence
[tex]\frac{a}{bc+1}\le\frac{2a}{S}[/tex]
In the same way [tex]\frac{b}{ca+1}\le\frac{2b}{S}[/tex] and [tex]\frac{c}{ab+1}\le\frac{2c}{S}[/tex]. Adding the three estimates:
[tex]\frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1}\le\frac{2\left(a+b+c\right)}{S}=2[/tex]
Together with the condition, the left side equals [tex]2[/tex] and all three estimates are equalities. One of the numbers is positive; by symmetry let [tex]a\gt 0[/tex]. Then equality in [tex]\frac{a}{bc+1}\le\frac{2a}{S}[/tex] means [tex]S=2\left(1+bc\right)[/tex], which requires equality in [tex]a\le 1+bc[/tex]; since [tex]a\le 1[/tex] and [tex]bc\ge 0[/tex], this forces [tex]a=1[/tex] and [tex]bc=0[/tex]. Therefore
[tex]S=2\left(1+0\right)=2[/tex]
Such numbers exist: for [tex]a=b=1[/tex], [tex]c=0[/tex] the left side is [tex]1+1+0=2[/tex]. The answer is [tex]a+b+c=2[/tex].