Proving Inequalities: Very Difficult Problems with Solutions

Solution:
Write each fraction as
[tex]\frac{a}{b^2+1}=a-\frac{ab^2}{b^2+1}[/tex]
Since [tex]b^2+1\ge 2b[/tex], for [tex]b\gt 0[/tex] we get [tex]\frac{ab^2}{b^2+1}\le\frac{ab^2}{2b}=\frac{ab}{2}[/tex], and for [tex]b=0[/tex] both sides are [tex]0[/tex]. Hence
[tex]\frac{a}{b^2+1}\ge a-\frac{ab}{2}[/tex]
and similarly [tex]\frac{b}{c^2+1}\ge b-\frac{bc}{2}[/tex] and [tex]\frac{c}{a^2+1}\ge c-\frac{ca}{2}[/tex]. Adding, we find that the left side is at least
[tex]a+b+c-\frac{ab+bc+ca}{2}=3-\frac{ab+bc+ca}{2}[/tex]
From [tex]\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge 0[/tex] we get [tex]a^2+b^2+c^2\ge ab+bc+ca[/tex], so
[tex]9=\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge 3\left(ab+bc+ca\right)\Rightarrow ab+bc+ca\le 3[/tex]
Therefore the left side is at least [tex]3-\frac{3}{2}=\frac{3}{2}[/tex], which proves the inequality. Equality holds for [tex]a=b=c=1[/tex].

Solution:
We bound each number by a linear expression in its cube. For every [tex]t\ge -1[/tex]
[tex]4t^3-3t+1=\left(t+1\right)\left(2t-1\right)^2\ge 0[/tex]
because [tex]t+1\ge 0[/tex] and a square is non-negative. Hence for every [tex]t\in\left[-1, 1\right][/tex]
[tex]t\le\frac{4t^3+1}{3}[/tex]
Apply this to each [tex]a_i[/tex] and add the nine inequalities:
[tex]a_1+a_2+\ldots+a_9\le\frac{4\left(a_1^3+a_2^3+\ldots+a_9^3\right)+9}{3}=\frac{0+9}{3}=3[/tex]
Equality requires every [tex]a_i[/tex] to be [tex]-1[/tex] or [tex]\frac{1}{2}[/tex]. Take [tex]a_1=-1[/tex] and [tex]a_2=a_3=\ldots=a_9=\frac{1}{2}[/tex]: all numbers lie in [tex]\left[-1, 1\right][/tex], the sum of the cubes is [tex]-1+8\times \frac{1}{8}=0[/tex] and the sum is [tex]-1+8\times \frac{1}{2}=3[/tex]. So the greatest value is [tex]3[/tex].

Solution:
We will use the inequality [tex]p^2+q^2+s^2\ge pq+qs+sp[/tex], which holds for all real [tex]p[/tex], [tex]q[/tex], [tex]s[/tex], because
[tex]p^2+q^2+s^2-\left(pq+qs+sp\right)=\frac{1}{2}\left[\left(p-q\right)^2+\left(q-s\right)^2+\left(s-p\right)^2\right]\ge 0[/tex]
Adding [tex]2\left(pq+qs+sp\right)[/tex] to both sides, we also get
[tex]\left(p+q+s\right)^2\ge 3\left(pq+qs+sp\right)[/tex]
Apply the first inequality to [tex]p=\frac{1}{a}[/tex], [tex]q=\frac{1}{b}[/tex], [tex]s=\frac{1}{c}[/tex] and use [tex]a+b+c=3[/tex]:
[tex]\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c}{abc}=\frac{3}{abc}[/tex]
So it suffices to prove that [tex]\frac{3}{abc}\ge a^2+b^2+c^2[/tex], i.e. [tex]abc\left(a^2+b^2+c^2\right)\le 3[/tex]. Let [tex]m=ab+bc+ca\gt 0[/tex]. Then
[tex]a^2+b^2+c^2=\left(a+b+c\right)^2-2m=9-2m[/tex]
in particular [tex]9-2m\gt 0[/tex]. Applying the second inequality to the numbers [tex]ab[/tex], [tex]bc[/tex], [tex]ca[/tex], we get
[tex]m^2=\left(ab+bc+ca\right)^2\ge 3\left(ab\times bc+bc\times ca+ca\times ab\right)=3abc\left(a+b+c\right)=9abc\Rightarrow abc\le\frac{m^2}{9}[/tex]
Therefore
[tex]abc\left(a^2+b^2+c^2\right)\le\frac{m\times m\times \left(9-2m\right)}{9}[/tex]
The numbers [tex]m[/tex], [tex]m[/tex], [tex]9-2m[/tex] are positive and their sum is [tex]9[/tex], so by the AM-GM inequality
[tex]m\times m\times \left(9-2m\right)\le\left(\frac{m+m+\left(9-2m\right)}{3}\right)^3=27[/tex]
Hence [tex]abc\left(a^2+b^2+c^2\right)\le\frac{27}{9}=3[/tex], which completes the proof. Equality holds for [tex]a=b=c=1[/tex].

Solution:
First we show that each of the three numbers lies in the interval [tex]\left(0 , 4\right)[/tex]. From the conditions, [tex]b+c=6-a[/tex] and
[tex]bc=9-a\left(b+c\right)=9-a\left(6-a\right)=\left(a-3\right)^2[/tex]
So [tex]b[/tex] and [tex]c[/tex] are the roots of the equation [tex]z^2-\left(6-a\right)z+\left(a-3\right)^2=0[/tex]. They are real and different, so its discriminant is positive:
[tex]D=\left(6-a\right)^2-4\left(a-3\right)^2=3a\left(4-a\right)\gt 0\Rightarrow 0\lt a\lt 4[/tex]
The conditions are symmetric, so in the same way [tex]0\lt b\lt 4[/tex] and [tex]0\lt c\lt 4[/tex]. Now let [tex]r=abc[/tex]. By Vieta's formulas, [tex]a[/tex], [tex]b[/tex], [tex]c[/tex] are the roots of the polynomial
[tex]p\left(t\right)=\left(t-a\right)\left(t-b\right)\left(t-c\right)=t^3-6t^2+9t-r[/tex]
From [tex]p\left(a\right)=0[/tex] we get [tex]r=a^3-6a^2+9a[/tex], hence
[tex]4-r=-\left(a^3-6a^2+9a-4\right)=\left(a-1\right)^2\left(4-a\right)\ge 0[/tex]
and in the same way [tex]4-r=\left(b-1\right)^2\left(4-b\right)=\left(c-1\right)^2\left(4-c\right)[/tex]. If [tex]r=4[/tex], then, since none of the numbers equals [tex]4[/tex], we get [tex]a=b=c=1[/tex], which contradicts [tex]a\lt b\lt c[/tex]. Hence [tex]r\lt 4[/tex]. Also [tex]r=abc\gt 0[/tex], because the three numbers are positive. Therefore
[tex]p\left(0\right)=-r\lt 0,\quad p\left(1\right)=4-r\gt 0,\quad p\left(3\right)=-r\lt 0,\quad p\left(4\right)=4-r\gt 0[/tex]
A polynomial that takes values of opposite signs at the ends of an interval has a root inside it, so [tex]p[/tex] has a root in each of the intervals [tex]\left(0 , 1\right)[/tex], [tex]\left(1 , 3\right)[/tex] and [tex]\left(3 , 4\right)[/tex]. These are three different roots of a cubic polynomial, so they are exactly [tex]a\lt b\lt c[/tex] in this order, which gives [tex]0\lt a\lt 1\lt b\lt 3\lt c\lt 4[/tex].

Solution:
We will prove the chain [tex]x^3+y^3\le x^2+y^2\le x+y\le 2[/tex]. The condition can be written as
[tex]x^2\left(1-x\right)\ge y^3\left(y-1\right)=-y^3\left(1-y\right)[/tex]
Adding [tex]y^2\left(1-y\right)[/tex] to both sides, we get
[tex]x^2\left(1-x\right)+y^2\left(1-y\right)\ge y^2\left(1-y\right)-y^3\left(1-y\right)=y^2\left(1-y\right)^2\ge 0[/tex]
i.e. [tex]x^3+y^3\le x^2+y^2[/tex]. Next, since [tex]x\ge 0[/tex] and [tex]y\ge 0[/tex],
[tex]\left(x+y\right)\left(x^3+y^3\right)-\left(x^2+y^2\right)^2=x^3y+xy^3-2x^2y^2=xy\left(x-y\right)^2\ge 0[/tex]
Together with the first inequality and [tex]x+y\ge 0[/tex] this gives
[tex]\left(x^2+y^2\right)^2\le\left(x+y\right)\left(x^3+y^3\right)\le\left(x+y\right)\left(x^2+y^2\right)[/tex]
If [tex]x=y=0[/tex], the claim is obvious. Otherwise [tex]x^2+y^2\gt 0[/tex], and dividing by it we get [tex]x^2+y^2\le x+y[/tex]. Finally, [tex]\left(x+y\right)^2\le 2\left(x^2+y^2\right)[/tex], since this is equivalent to [tex]\left(x-y\right)^2\ge 0[/tex]; hence
[tex]\left(x+y\right)^2\le 2\left(x^2+y^2\right)\le 2\left(x+y\right)[/tex]
Dividing by [tex]x+y\gt 0[/tex], we get [tex]x+y\le 2[/tex]. Therefore
[tex]x^3+y^3\le x^2+y^2\le x+y\le 2[/tex]
Equality holds for [tex]x=y=1[/tex].

Solution:
Necessity. Put [tex]a=b=c=s\gt 0[/tex]; the inequality becomes [tex]3rs^2+\frac{3\left(3-r\right)}{s}\ge 9[/tex]. Multiplying by [tex]\frac{s}{3}\gt 0[/tex] and moving everything to the left side, we see that for all [tex]s\gt 0[/tex] we need
[tex]rs^3-3s+3-r=r\left(s^3-1\right)-3\left(s-1\right)=\left(s-1\right)\left(r\left(s^2+s+1\right)-3\right)\ge 0[/tex]
Note that [tex]s^2+s+1[/tex] is close to [tex]3[/tex] when [tex]s[/tex] is close to [tex]1[/tex].
If [tex]r\gt 1[/tex], then [tex]\frac{3}{r}\lt 3[/tex], and we can choose [tex]s\lt 1[/tex] so close to [tex]1[/tex] that [tex]s^2+s+1\gt\frac{3}{r}[/tex]; then the first factor is negative and the second one is positive, so the product is negative, a contradiction.
If [tex]r\le 0[/tex], the second factor is always negative, and every [tex]s\gt 1[/tex] gives a negative product.
If [tex]0\lt r\lt 1[/tex], then [tex]\frac{3}{r}\gt 3[/tex], and we choose [tex]s\gt 1[/tex] so close to [tex]1[/tex] that [tex]s^2+s+1\lt\frac{3}{r}[/tex]; the product is again negative. Hence only [tex]r=1[/tex] is possible.
Sufficiency. For [tex]r=1[/tex] we have to prove that
[tex]ab+bc+ca+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge 9[/tex]
By the AM-GM inequality for three positive numbers,
[tex]ab+\frac{1}{a}+\frac{1}{b}\ge 3\sqrt[3]{ab\times \frac{1}{a}\times \frac{1}{b}}=3[/tex]
and similarly [tex]bc+\frac{1}{b}+\frac{1}{c}\ge 3[/tex] and [tex]ca+\frac{1}{c}+\frac{1}{a}\ge 3[/tex]. Adding these three inequalities gives exactly the required one. Therefore the only such number is [tex]r=1[/tex].

Solution:
Put [tex]u=2a[/tex] and [tex]v=c[/tex]; then [tex]u\gt 0[/tex], [tex]v\gt 0[/tex], [tex]u\ne v[/tex], and the condition, after rationalising its right side, reads
[tex]u-v=\sqrt{b+\sqrt{u}}-\sqrt{b+\sqrt{v}}=\frac{\sqrt{u}-\sqrt{v}}{\sqrt{b+\sqrt{u}}+\sqrt{b+\sqrt{v}}}[/tex]
On the other hand, [tex]u-v=\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)[/tex]. Since [tex]u\ne v[/tex], we have [tex]\sqrt{u}-\sqrt{v}\ne 0[/tex], and dividing by it we get
[tex]\sqrt{u}+\sqrt{v}=\frac{1}{\sqrt{b+\sqrt{u}}+\sqrt{b+\sqrt{v}}}[/tex]
Because [tex]b\gt 0[/tex], we have [tex]\sqrt{b+\sqrt{u}}\gt\sqrt[4]{u}[/tex] and [tex]\sqrt{b+\sqrt{v}}\gt\sqrt[4]{v}[/tex], therefore
[tex]\left(\sqrt{u}+\sqrt{v}\right)\left(\sqrt[4]{u}+\sqrt[4]{v}\right)\lt 1[/tex]
By the AM-GM inequality, [tex]\sqrt{u}+\sqrt{v}\ge 2\sqrt[4]{uv}[/tex] and [tex]\sqrt[4]{u}+\sqrt[4]{v}\ge 2\sqrt[8]{uv}[/tex], so
[tex]4\sqrt[8]{\left(uv\right)^3}\lt 1\Rightarrow\left(uv\right)^3\lt\frac{1}{4^8}=\frac{1}{2^{16}}[/tex]
Since [tex]uv=2ac[/tex], this means
[tex]8\left(ac\right)^3\lt\frac{1}{2^{16}}\Rightarrow\left(ac\right)^3\lt\frac{1}{2^{19}}[/tex]
Finally, [tex]2^{19}=524288\gt 512000=80^3[/tex], so [tex]\left(ac\right)^3\lt\frac{1}{80^3}[/tex], hence [tex]ac\lt\frac{1}{80}[/tex], i.e. [tex]80ac\lt 1[/tex].

Solution:
For [tex]a=b=1[/tex], [tex]c=0[/tex] the inequality becomes [tex]8\ge 2k[/tex], so [tex]k\le 4[/tex]. We will show that [tex]k=4[/tex] works. Denote
[tex]T=a^2b+a^2c+b^2a+b^2c+c^2a+c^2b[/tex]
Then
[tex]\left(a+b+c\right)^3=a^3+b^3+c^3+3T+6abc,\quad\left(a+b+c\right)\left(ab+bc+ca\right)=T+3abc[/tex]
For [tex]k=4[/tex] the right side equals [tex]4T+12abc-9abc=4T+3abc[/tex], so the inequality is equivalent to
[tex]a^3+b^3+c^3-T+3abc\ge 0[/tex]
Expanding shows that the left side is exactly the expression from Schur's inequality:
[tex]a^3+b^3+c^3-T+3abc=a\left(a-b\right)\left(a-c\right)+b\left(b-a\right)\left(b-c\right)+c\left(c-a\right)\left(c-b\right)[/tex]
We prove that it is non-negative. The expression is symmetric, so we may assume [tex]a\ge b\ge c\ge 0[/tex]. Then it equals
[tex]\left(a-b\right)\left[a\left(a-c\right)-b\left(b-c\right)\right]+c\left(a-c\right)\left(b-c\right)[/tex]
Here [tex]a-b\ge 0[/tex]; [tex]a\left(a-c\right)\ge b\left(b-c\right)[/tex], because [tex]a\ge b\ge 0[/tex] and [tex]a-c\ge b-c\ge 0[/tex]; and [tex]c\left(a-c\right)\left(b-c\right)\ge 0[/tex]. So the expression is non-negative and [tex]k=4[/tex] satisfies the inequality for all non-negative [tex]a[/tex], [tex]b[/tex], [tex]c[/tex], while every [tex]k\gt 4[/tex] fails for [tex]a=b=1[/tex], [tex]c=0[/tex]. The answer is [tex]k=4[/tex].

Solution:
Rewrite each term as follows (bringing the right side to a common denominator gives the numerator [tex]4x^2+4x+1-\left(2x^2+1\right)=2x^2+4x[/tex]):
[tex]\frac{x\left(x+2\right)}{2x^2+1}=\frac{\left(2x+1\right)^2}{2\left(2x^2+1\right)}-\frac{1}{2}[/tex]
So the inequality is equivalent to
[tex]\frac{\left(2x+1\right)^2}{2x^2+1}+\frac{\left(2y+1\right)^2}{2y^2+1}+\frac{\left(2z+1\right)^2}{2z^2+1}\ge 3[/tex]
Among three numbers some two have the same sign or one of them is zero, so some two of them have a non-negative product; since the inequality is symmetric, we may assume [tex]xy\ge 0[/tex]. Then, because [tex]x+y=-z[/tex],
[tex]x^2+y^2\le x^2+2xy+y^2=\left(x+y\right)^2=z^2[/tex]
By the Cauchy-Schwarz inequality in the form [tex]\frac{p^2}{m}+\frac{q^2}{n}\ge\frac{\left(p+q\right)^2}{m+n}[/tex] for [tex]m\gt 0[/tex], [tex]n\gt 0[/tex]:
[tex]\frac{\left(2x+1\right)^2}{2x^2+1}+\frac{\left(2y+1\right)^2}{2y^2+1}\ge\frac{\left(2x+2y+2\right)^2}{2x^2+2y^2+2}=\frac{2\left(x+y+1\right)^2}{x^2+y^2+1}\ge\frac{2\left(1-z\right)^2}{z^2+1}[/tex]
In the last step we used [tex]\left(x+y+1\right)^2=\left(1-z\right)^2[/tex] and [tex]0\lt x^2+y^2+1\le z^2+1[/tex]. It remains to prove the one-variable inequality
[tex]\frac{2\left(1-z\right)^2}{z^2+1}+\frac{\left(2z+1\right)^2}{2z^2+1}\ge 3[/tex]
Multiply by the positive number [tex]\left(z^2+1\right)\left(2z^2+1\right)[/tex] and move everything to the left side:
[tex]2\left(1-z\right)^2\left(2z^2+1\right)+\left(2z+1\right)^2\left(z^2+1\right)-3\left(z^2+1\right)\left(2z^2+1\right)[/tex]
[tex]=\left(4z^4-8z^3+6z^2-4z+2\right)+\left(4z^4+4z^3+5z^2+4z+1\right)-\left(6z^4+9z^2+3\right)[/tex]
[tex]=2z^4-4z^3+2z^2=2z^2\left(z-1\right)^2\ge 0[/tex]
This completes the proof. Equality holds, for example, for [tex]x=y=z=0[/tex] and for [tex]x=y=-\frac{1}{2}[/tex], [tex]z=1[/tex].