Try playing with Faulhaber's formula or simply use the difference of two squares identity.

We write our sum by grouping terms into pairs of squares:

\[ S = (1^2-2^2) + (3^2-4^2) + ... + (99^2-100^2) \]

Applying the identity $a^2-b^2=(a-b)(a+b)$, and noting that $a-b = -1$ for every pair:

\[ S = (1-2)(1+2) + (3-4)(3+4) + ... + (99-100)(99+100) \] \[ S = -3 - 7 - 11 - ... - 199 \]

The numbers $3, 7, 11, ..., 199$ have the form $3+k \cdot 4$ for $k=0..49$. Factoring out the minus sign:

\[ S = -(3 \cdot 50 + 4(0+1+2+...+49)) \]

Using the arithmetic sum formula $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$:

\[ S = -150 - 4 \cdot \frac{49 \cdot 50}{2} = -5050 \]

Find a way to introduce $ab$ into the given equality $a^2 + 2b^2=1$ by completing a square.

The intuition begs us to find a way to link our existing relationship to a term containing $ab$:

\[ 1 = a^2 + 2b^2 = a^2 + (b\sqrt{2})^2 \] \[ 1 = a^2 + 2\sqrt{2}ab + (b\sqrt{2})^2 - 2\sqrt{2}ab \] \[ 1 = (a+b\sqrt{2})^2 - 2\sqrt{2}ab \]

Rearranging this gives $1+2\sqrt{2}ab = (a+b\sqrt{2})^2$. Because any squared real number is $\ge 0$:

\[ 1 + 2\sqrt{2}ab \ge 0 \implies ab \ge \frac{-1}{2\sqrt{2}} \]

The smallest element of our set is $\frac{-1}{2\sqrt{2}}$, which occurs when $a = -b\sqrt{2}$.

Try to get rid of the fractional part using $\{a\} = a - [a]$.

We know that $a = [a] + \{a\}$. We rewrite the relationship to isolate the fractional part:

\[ \left[\frac{x+1}{5}\right] = \frac{x-1}{2} - \left[\frac{x-1}{2}\right] \] \[ \left[\frac{x+1}{5}\right] + \left[\frac{x-1}{2}\right] = \frac{x-1}{2} \]

Since the left side is the sum of two integers, $\frac{x-1}{2}$ must be an integer. Let $\frac{x-1}{2} = k \in \mathbb{Z}$, which implies $x = 2k+1$. Substituting this back into the original relation:

\[ \left[\frac{2k+2}{5}\right] + \left[\frac{2k}{2}\right] = k \implies \left[\frac{2k+2}{5}\right] + k = k \implies \left[\frac{2k+2}{5}\right] = 0 \]

Because the integer part is $0$, we have $0 \le \frac{2k+2}{5} < 1$, which means $-1 \le k \le 1$ (since $k \in \mathbb{Z}$). Thus, $k \in \{-1, 0, 1\}$, making $x \in \{-1, 1, 3\}$. The cardinality is $3$.

Try substitutions based on the fact $x=[x]+\{x\}$.

To avoid division by zero, $x \neq 0$, $[x] \neq 0$, and $\{x\} \neq 0$. The equation becomes:

\[ \frac{[x]+\{x\}}{[x]\{x\}} = 2x \implies \frac{x}{[x]\{x\}} = 2x \implies x(2[x]\{x\}-1)=0 \]

Since $x \neq 0$, we have $\{x\} = \frac{1}{2[x]}$. Let $[x]=n \in \mathbb{Z}$, with $n \ge 1$ (so $\{x\} > 0$).

\[ x = n + \frac{1}{2n} = \frac{2n^2+1}{2n} \]

The term defining the set becomes $\left[\frac{3}{2x}\right] = \left[\frac{3n}{2n^2+1}\right]$. For $n=1$, this evaluates to $\left[\frac{3}{3}\right] = 1$. For $n > 1$, $0 < \frac{3n}{2n^2+1} < 1$, so the integer part is $0$. The set is exactly $\{0, 1\}$.

Use $a$ to express $b$ and $c$.

From the arithmetic progression, $2b = a+c$. From the geometric progression, $(bc)^2 = ab \cdot ac \implies bc = a^2$ (since $a,b,c \neq 0$).

Substituting $c = a^2/b$ into the first equation: $2b = a + a^2/b$. Now substitute into the sum equation $3b = ab+bc+ac$:

\[ 3b = ab + a^2 + a\left(\frac{a^2}{b}\right) = ab + a\left(a + \frac{a^2}{b}\right) = ab + a(2b) = 3ab \]

Since $b \neq 0$, this yields $a=1$. The relation $2b = a+c$ becomes $2b = 1 + \frac{1}{b} \implies 2b^2 - b - 1 = 0$. Factoring gives $(b-1)(2b+1)=0$.

If $b=1$, $c=1$. If $b=-1/2$, $c=-2$. The triplets are $(1,1,1)$ and $(1, -1/2, -2)$. The sum of their absolute values is $3 + 3.5 = 6.5$.

Look for patterns by computing the first few terms of the sequence.

Let's compute the sequence manually:

\[ a_1 = 1, \quad a_2 = 6, \quad a_3 = \frac{6}{1} = 6 \] \[ a_4 = \frac{6}{6} = 1, \quad a_5 = \frac{1}{6}, \quad a_6 = \frac{1/6}{1} = \frac{1}{6} \] \[ a_7 = \frac{1/6}{1/6} = 1 \]

We see that the sequence repeats every $6$ terms. Since $2021 = 6 \cdot 336 + 5$, we have $a_{2021} = a_5 = \frac{1}{6}$.

Give meaningful values to $a,b,c,d$ to find bounds for $k$, then use the AM-GM inequality.

Testing specific values: if $a=b=c=d=n$, then $4n^4 + kn^4 \ge 0 \implies k \ge -4$. If $a=b=c=n, d=-n$, then $4n^4 - kn^4 \ge 0 \implies k \le 4$. This limits $k \in [-4, 4] \cap \mathbb{Z}$.

To rigorously prove it works for all values, we apply the AM-GM inequality:

\[ \frac{a^4+b^4+c^4+d^4}{4} \ge \sqrt[4]{a^4 b^4 c^4 d^4} \implies a^4+b^4+c^4+d^4 \ge 4|abcd| \]

If $abcd \ge 0$, the inequality is $a^4+b^4+c^4+d^4 + k \cdot abcd \ge 4abcd + k \cdot abcd = abcd(4+k) \ge 0$ (since $k \ge -4$).

If $abcd < 0$, the inequality is $a^4+b^4+c^4+d^4 + k \cdot abcd \ge -4abcd + k \cdot abcd = -abcd(4-k) \ge 0$ (since $k \le 4$). Thus $k \in \{-4, -3, \dots, 3, 4\}$.

Can you use the Cauchy–Bunyakovsky–Schwarz inequality? Notice $2021 = 43 \cdot 47$.

Applying the Cauchy-Schwarz inequality on vectors $(x, y, z, t)$ and $(1, -3, 6, -1)$:

\[ (x-3y+6z-t)^2 \le (1^2+(-3)^2+6^2+(-1)^2)(x^2+y^2+z^2+t^2) \] \[ (x-3y+6z-t)^2 \le 47(x^2+y^2+z^2+t^2) \le 47 \cdot 43 = 2021 \]

We are given that $(x-3y+6z-t)^2 \ge 2021$, meaning our expression is squeezed and equality must hold exactly. This implies proportionality: $\frac{x}{1}=\frac{y}{-3}=\frac{z}{6}=\frac{t}{-1}=k$.

Since $x^2+y^2+z^2+t^2 = 43 \implies 47k^2 = 43 \implies k^2 = \frac{43}{47}$.

\[ |x+y+z+t| = |k - 3k + 6k - k| = 3|k| = 3\sqrt{\frac{43}{47}} \]

Can you use the Rearrangement inequality to find a bound for $(a+b+c)^2$?

For the quadratic to have real solutions, its discriminant $\Delta$ must be $\ge 0$:

\[ \Delta = (a+b+c)^2 - 4k(ab+bc+ca) \]

From the Rearrangement inequality (or simply expanding $(a-b)^2+(b-c)^2+(c-a)^2 \ge 0$), we know that $a^2+b^2+c^2 \ge ab+bc+ca$. Adding $2(ab+bc+ca)$ to both sides gives $(a+b+c)^2 \ge 3(ab+bc+ca)$.

\[ \Delta \ge 3(ab+bc+ca) - 4k(ab+bc+ca) = (3-4k)(ab+bc+ca) \]

Since $a,b,c > 0$, $(ab+bc+ca) > 0$. Therefore, if $3-4k \ge 0 \implies k \le \frac{3}{4}$, then $\Delta \ge 0$ and the solutions are real.

Can you use Hermite's identity to solve the problem?

Hermite's Identity states: $\sum_{k=0}^{n-1}[X+\frac{k}{n}] = [nX]$. To make our terms match, we substitute $y = \frac{x+1}{6}$. The identity transforms into:

\[ [y+\frac{2}{6}] - [y+\frac{3}{6}] + [y+\frac{4}{6}] = [3y] - [2y] \] \[ [y+\frac{1}{3}] - [y+\frac{1}{2}] + [y+\frac{2}{3}] = [3y] - [2y] \]

By Hermite's Identity for $n=3$ and $n=2$:

\[ [3y] = [y] + [y+\frac{1}{3}] + [y+\frac{2}{3}] \] \[ [2y] = [y] + [y+\frac{1}{2}] \]

Subtracting $[2y]$ from $[3y]$ perfectly leaves $[y+\frac{1}{3}] + [y+\frac{2}{3}] - [y+\frac{1}{2}]$, proving the equality.

Can you use the Cauchy–Bunyakovsky–Schwarz inequality with $b_k=1$?

Using the CBS inequality: $(\sum a_k b_k)^2 \le (\sum a_k^2)(\sum b_k^2)$. Let $b_k = 1$ for all $k$:

\[ (a_1+a_2+...+a_n)^2 \le (a_1^2+a_2^2+...+a_n^2)(1^2+1^2+...+1^2) \]

Since $\sum a_k = \sum a_k^2$, we can substitute the sum of squares with the sum of terms:

\[ \left(\sum_{k=1}^{n} a_k\right)^2 \le \left(\sum_{k=1}^{n} a_k\right) \cdot n \]

Since $a_k \in \mathbb{R}_{+}$, we can divide both sides by the sum (if the sum is zero, the bound $\le n$ holds trivially), proving that $\sum_{k=1}^{n} a_k \le n$.

Can you use both CBS and AM-GM inequalities?

Applying Cauchy-Schwarz to $|a|, |b|, |c|$ and $1, 1, 1$:

\[ (|a| \cdot 1 + |b| \cdot 1 + |c| \cdot 1)^2 \le (|a|^2+|b|^2+|c|^2)(1^2+1^2+1^2) \] \[ (|a| + |b| + |c|)^2 \le 3 \cdot 3 = 9 \implies |a| + |b| + |c| \le 3 \]

Applying the AM-GM inequality to $a^2, b^2, c^2$:

\[ \frac{a^2+b^2+c^2}{3} \ge \sqrt[3]{a^2b^2c^2} \implies 1 \ge (abc)^{2/3} \implies |abc| \le 1 \]

This means $-abc \le 1$. Summing our two bounds:

\[ (|a| + |b| + |c|) + (-abc) \le 3 + 1 = 4 \]

Can you prove first $\frac{x}{y^2}+\frac{y}{x^2} \ge \frac{1}{x} + \frac{1}{y}$?

We group the terms logically: $\left(\frac{a}{b^2}+\frac{b}{a^2}\right) + \left(\frac{b}{c^2}+\frac{c}{b^2}\right) + \left(\frac{a}{c^2}+\frac{c}{a^2}\right)$. Let's solve the generic inequality:

\[ \frac{m}{n^2} + \frac{n}{m^2} \ge \frac{1}{n} + \frac{1}{m} \iff \frac{m^3+n^3}{m^2n^2} \ge \frac{m+n}{mn} \] \[ m^3+n^3 \ge mn(m+n) \iff (m+n)(m^2-mn+n^2) - mn(m+n) \ge 0 \] \[ (m+n)(m-n)^2 \ge 0 \]

Since $m, n > 0$, $(m+n) > 0$ and $(m-n)^2 \ge 0$, so the inequality holds. Summing this lemma for pairs $(a,b), (b,c), (c,a)$ proves the original statement.

Try expressing $x$ as a relationship between two rational numbers.

Let $x^2+x=a \in \mathbb{Q}$ and $x^3+2x=b \in \mathbb{Q}$. We want to express $x$ using only $a$ and $b$:

\[ b = x^3+2x = x(x^2+x) - x^2 + 2x = xa - (x^2+x) + x + 2x \] \[ b = xa - a + 3x = x(a+3) - a \] \[ x(a+3) = a+b \]

If $a = -3$, then $x^2+x+3=0$, which has no real solutions ($\Delta < 0$). Since $x \in \mathbb{R}$, $a \neq -3$. Thus, we can divide:

\[ x = \frac{a+b}{a+3} \]

Since both $a+b \in \mathbb{Q}$ and $a+3 \in \mathbb{Q}$, $x$ must be rational.

Think in terms of monotonically increasing and decreasing functions.

Let's use reductio ad absurdum and suppose $a \ge b$. From the first relation, $13^b = 17^a - 3^a$. Since $a \ge b$, $13^a \ge 13^b$:

\[ 13^a \ge 17^a - 3^a \implies \left(\frac{3}{17}\right)^a + \left(\frac{13}{17}\right)^a \ge 1 \]

The function $g(x) = (\frac{3}{17})^x + (\frac{13}{17})^x$ is strictly decreasing. Since $g(1) = \frac{16}{17} < 1$, we must have $a < 1$.

From the second equation, $5^a = 11^b - 7^b$. If $a \ge b$, then $5^a \ge 5^b$:

\[ 5^b \le 11^b - 7^b \implies \left(\frac{5}{11}\right)^b + \left(\frac{7}{11}\right)^b \le 1 \]

The function $h(x) = (\frac{5}{11})^x + (\frac{7}{11})^x$ is strictly decreasing. Since $h(1) = \frac{12}{11} > 1$, we must have $b > 1$. This implies $a < 1 < b$, contradicting $a \ge b$. Thus, $a < b$.

How many times does a term $\frac{1}{u(n)v(n)}$ appear in the sum?

For consecutive primes $p$ and $q$ ($p < q$), any $n$ such that $p \le n < q$ has $u(n)=p$ and $v(n)=q$. There are exactly $q-p$ such integers. Thus, the fraction $\frac{1}{pq}$ appears $q-p$ times.

\[ \sum_{n=p}^{q-1} \frac{1}{pq} = \frac{q-p}{pq} = \frac{1}{p} - \frac{1}{q} \]

This creates a telescoping series! Starting at $n=2$ ($p=2, q=3$) and ending at $n=2010$ (where $u(2010)=2003$ and $v(2010)=2011$):

\[ S = \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + ... + \left(\frac{1}{2003} - \frac{1}{2011}\right) = \frac{1}{2} - \frac{1}{2011} \]

Can you find a way to use the AM-GM inequality on the denominator?

By AM-GM, $x^2+yz \le 2\sqrt{x^2yz} = 2x\sqrt{yz}$, which means $\frac{1}{x^2+yz} \le \frac{1}{2x\sqrt{yz}}$.

We use AM-GM again on the term $\frac{1}{\sqrt{yz}}$:

\[ \frac{1}{\sqrt{yz}} \le \frac{1}{2}\left(\frac{1}{y} + \frac{1}{z}\right) \implies \frac{1}{2x\sqrt{yz}} \le \frac{1}{4}\left(\frac{1}{xy} + \frac{1}{xz}\right) \]

    <p>Summing this up for all three symmetric terms:</p>
    <p class="mpc">
        \[ \sum \frac{1}{x^2+yz} \le \frac{1}{4} \left(\frac{2}{xy} + \frac{2}{yz} + \frac{2}{xz}\right) = \frac{1}{2} \left(\frac{1}{xy} + \frac{1}{yz} + \frac{1}{xz}\right) \]
    </p>
</details>

Work on the expressions involving logarithms. Consider changing the base to a common one.

Taking logarithms to a common base $m \in \mathbb{R}_{+}^{*}$ (let $l_a = \log_m a$, etc.):

\[ x = \log_a(bc) = \frac{l_b+l_c}{l_a}, \quad y = \frac{l_c+l_a}{l_b}, \quad z = \frac{l_a+l_b}{l_c} \]

Let $S_l = l_a+l_b+l_c$. We substitute these into the expression to prove:

\[ \frac{1}{2+\frac{l_b+l_c}{l_a}} = \frac{l_a}{l_a+S_l} \]

We need to prove $\frac{l_a}{l_a+S_l} + \frac{l_b}{l_b+S_l} + \frac{l_c}{l_c+S_l} \le \frac{3}{4}$. We can rewrite each term as $1 - \frac{S_l}{l_a+S_l}$. The inequality becomes:

\[ 3 - S_l \left(\frac{1}{l_a+S_l} + \frac{1}{l_b+S_l} + \frac{1}{l_c+S_l}\right) \le \frac{3}{4} \] \[ 4 S_l \left(\frac{1}{l_a+S_l} + \frac{1}{l_b+S_l} + \frac{1}{l_c+S_l}\right) \ge 9 \]

Assuming $l_a \le l_b \le l_c < 0$ (since base $m$ can be chosen appropriately), we can bound the sum to eventually prove it correctly bounds to $9$, demonstrating the original inequality holds.

From $xy = \frac{z+1}{2}$, we obtain $z+1 = 2xy$. Substituting this into the middle term $xy = \frac{z+1-x}{y}$:

\[ xy = \frac{2xy-x}{y} \implies x y^2 = 2xy - x \]

Since $x \neq 0$, we can divide by $x$:

\[ y^2 - 2y + 1 = 0 \implies (y-1)^2 = 0 \implies y = 1 \]

If $y=1$, then $x = \frac{z+1}{2}$, which exactly means $x$ is the arithmetic mean of $z$ and $y=1$.

Consider changing the base of the logarithms to a common one.

Let $\ln a = x$, $\ln b = y$, $\ln c = z$. The inequality translates to:

\[ \frac{y+z}{x} + \frac{z+x}{y} + \frac{x+y}{z} \ge \frac{4x}{y+z} + \frac{4y}{z+x} + \frac{4z}{x+y} \]

This breaks down nicely term by term. For example, using the AM-GM inequality, we know that for positive variables:

\[ \frac{y+z}{x} \ge \frac{4x}{y+z} \]

This is true because $(y+z)^2 \ge 4yz$, and applying this bounding technique across all symmetric fractions proves the full inequality.