Olympiad MethodsDifficulty 70-1001 tagged problems

Holder's Inequality

Holder's Inequality is the habit of generalizing Cauchy-Schwarz to bound products and sums in harder inequality problems. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$\sum_{i=1}^n |a_i b_i| \;\le\; \left(\sum_{i=1}^n |a_i|^p\right)^{1/p}\left(\sum_{i=1}^n |b_i|^q\right)^{1/q}, \qquad \frac{1}{p} + \frac{1}{q} = 1,\ p,q>1.$$

Hölder's inequality bounds a sum of products by the product of $L^p$ and $L^q$ norms whenever the exponents are conjugate ($\tfrac1p+\tfrac1q=1$); taking $p=q=2$ recovers Cauchy–Schwarz.

Why It Works

1
The key pointwise tool is Young's inequality: for $u,v \ge 0$ and conjugate exponents $\frac1p+\frac1q = 1$, $uv \le \frac{u^p}{p} + \frac{v^q}{q}$ (it follows from concavity of $\log$).
2
Normalize: let $A = \left(\sum |a_i|^p\right)^{1/p}$ and $B = \left(\sum |b_i|^q\right)^{1/q}$, and set $u_i = |a_i|/A$, $v_i = |b_i|/B$, so that $\sum u_i^p = \sum v_i^q = 1$.
3
Apply Young termwise: $\sum u_i v_i \le \sum\!\left(\frac{u_i^p}{p} + \frac{v_i^q}{q}\right) = \frac{1}{p} + \frac{1}{q} = 1$.
4
Unnormalizing ($u_iv_i = |a_ib_i|/(AB)$) gives $\sum |a_i b_i| \le AB$, which is the stated inequality; equality holds when $|a_i|^p$ and $|b_i|^q$ are proportional.

Worked Examples

Example 1

For positive reals $a,b,c$ with $a+b+c = 1$, find the minimum of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.

1
Use the $p=q=2$ case (Cauchy–Schwarz, a special case of Hölder) in Engel/sum form, or apply Hölder directly: $\left(\sum \frac{1}{a}\right)\left(\sum a\right) \ge \left(\sum 1\right)^2$ by Cauchy–Schwarz.
2
Substitute $\sum a = 1$ and $\sum 1 = 3$ (three terms): $\left(\frac1a+\frac1b+\frac1c\right)\cdot 1 \ge 3^2 = 9$.
3
Hence $\frac1a + \frac1b + \frac1c \ge 9$.
4
Equality requires $a=b=c$; with $a+b+c=1$ this gives $a=b=c=\frac13$, where $\frac1a+\frac1b+\frac1c = 9$.

Answer:

Minimum value $9$, attained at $a=b=c=\frac{1}{3}$.

Contest level

For positive reals $a, b, c$, prove $(a^3 + b^3 + c^3)(1+1+1)(1+1+1) \ge (a + b + c)^3$, i.e. $9(a^3+b^3+c^3) \ge (a+b+c)^3$.

1
Apply Hölder with three sequences and equal exponents $3,3,3$ (since $\tfrac13+\tfrac13+\tfrac13 = 1$): $\sum a\cdot 1\cdot 1 \le \left(\sum a^3\right)^{1/3}\left(\sum 1^3\right)^{1/3}\left(\sum 1^3\right)^{1/3}$.
2
The right side is $\left(\sum a^3\right)^{1/3}\cdot 3^{1/3}\cdot 3^{1/3}$ and the left side is $a+b+c$.
3
Cube both sides: $(a+b+c)^3 \le \left(\sum a^3\right)\cdot 3\cdot 3 = 9(a^3+b^3+c^3)$.
4
Equality holds when $a=b=c$. (This is exactly the power-mean inequality $M_1 \le M_3$.)

Answer:

$9(a^3+b^3+c^3) \ge (a+b+c)^3$, equality iff $a=b=c$.

Olympiad / Challenge

For positive reals $a, b, c$, prove $\dfrac{a}{\sqrt{a^2+8bc}} + \dfrac{b}{\sqrt{b^2+8ca}} + \dfrac{c}{\sqrt{c^2+8ab}} \ge 1$.

1
Apply the three-factor Hölder with equal weights $\tfrac13,\tfrac13,\tfrac13$ (the $(3,3,3)$ form, since $\tfrac13+\tfrac13+\tfrac13 = 1$) to the three sums $\sum \tfrac{a}{\sqrt{a^2+8bc}}$, $\sum \tfrac{a}{\sqrt{a^2+8bc}}$, and $\sum a(a^2+8bc)$: $\Bigl(\sum \tfrac{a}{\sqrt{a^2+8bc}}\Bigr)\Bigl(\sum \tfrac{a}{\sqrt{a^2+8bc}}\Bigr)\Bigl(\sum a(a^2+8bc)\Bigr) \ge \Bigl(\sum \sqrt[3]{\tfrac{a}{\sqrt{a^2+8bc}}\cdot\tfrac{a}{\sqrt{a^2+8bc}}\cdot a(a^2+8bc)}\Bigr)^3$. Inside the cube root the radicals cancel, $\tfrac{a^2}{a^2+8bc}\cdot a(a^2+8bc) = a^3$, so each summand is $\sqrt[3]{a^3} = a$. Writing $S = \sum \tfrac{a}{\sqrt{a^2+8bc}}$ this is $S^2\cdot\bigl[\sum a(a^2+8bc)\bigr] \ge (a+b+c)^3$.
2
So it suffices to show $(a+b+c)^3 \ge \sum a(a^2+8bc) = a^3+b^3+c^3 + 24abc$.
3
Expand $(a+b+c)^3 = a^3+b^3+c^3 + 3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) + 6abc$. The needed inequality reduces to $3(a^2b+a^2c+\cdots+c^2b) + 6abc \ge 24abc$, i.e. $a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \ge 6abc$.
4
That last step is AM–GM on the six terms: their geometric mean is $\sqrt[6]{a^6b^6c^6} = abc$, so the sum is $\ge 6abc$. Hence $S^2 \ge \dfrac{(a+b+c)^3}{\sum a(a^2+8bc)} \ge 1$, giving $S \ge 1$. Equality at $a=b=c$.

Answer:

$\displaystyle\sum \frac{a}{\sqrt{a^2+8bc}} \ge 1$, equality iff $a=b=c$.

Going Deeper

Hölder generalizes to any number of sequences with exponents summing reciprocally to $1$ ($\sum \tfrac{1}{p_k}=1$); the $L^p$-norm integral form $\int |fg| \le \|f\|_p\|g\|_q$ is the analyst's version, and $p=q=2$ is Cauchy–Schwarz.
The 'Hölder for fractions' template — $\bigl(\sum \tfrac{x_i}{y_i}\bigr)\cdot(\cdots) \ge (\sum \text{something})^k$ — is the standard finishing move on hard IMO/shortlist inequalities like the $\sqrt{a^2+8bc}$ problem above.
Pitfall: the exponents must be conjugate ($\tfrac1p+\tfrac1q=1$, both $>1$) and you must keep $a_i, b_i \ge 0$ (or take absolute values). Using $p=q=2$ when the natural pairing is $p=3,\,q=\tfrac32$ gives a weaker, often useless bound.

Spot the Signal

Look for problems where the key step is generalizing Cauchy-Schwarz to bound products and sums in harder inequality problems.
You can describe the hard part as generalizing Cauchy-Schwarz to bound products and sums in harder inequality problems, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the olympiad structure that calls for holder's inequality, then rewrite the givens around it.
Name the relation that makes Holder's Inequality legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Holder's Inequality just because the surface looks familiar; verify the required condition first.
Applying Holder's Inequality because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let holder's inequality reveal the structural theorem; then compute only what remains.

Try it on:

Practice a contest problem where the key step is generalizing Cauchy-Schwarz to bound products and sums in harder inequality problems.

Back to techniques

Olympiad MethodsDifficulty 70-1001 tagged problems

Holder's Inequality

Practice this technique Match my level

The Key Result

$$\sum_{i=1}^n |a_i b_i| \;\le\; \left(\sum_{i=1}^n |a_i|^p\right)^{1/p}\left(\sum_{i=1}^n |b_i|^q\right)^{1/q}, \qquad \frac{1}{p} + \frac{1}{q} = 1,\ p,q>1.$$

Hölder's inequality bounds a sum of products by the product of $L^p$ and $L^q$ norms whenever the exponents are conjugate ($\tfrac1p+\tfrac1q=1$); taking $p=q=2$ recovers Cauchy–Schwarz.

Why It Works

1
The key pointwise tool is Young's inequality: for $u,v \ge 0$ and conjugate exponents $\frac1p+\frac1q = 1$, $uv \le \frac{u^p}{p} + \frac{v^q}{q}$ (it follows from concavity of $\log$).
2
Normalize: let $A = \left(\sum |a_i|^p\right)^{1/p}$ and $B = \left(\sum |b_i|^q\right)^{1/q}$, and set $u_i = |a_i|/A$, $v_i = |b_i|/B$, so that $\sum u_i^p = \sum v_i^q = 1$.
3
Apply Young termwise: $\sum u_i v_i \le \sum\!\left(\frac{u_i^p}{p} + \frac{v_i^q}{q}\right) = \frac{1}{p} + \frac{1}{q} = 1$.
4
Unnormalizing ($u_iv_i = |a_ib_i|/(AB)$) gives $\sum |a_i b_i| \le AB$, which is the stated inequality; equality holds when $|a_i|^p$ and $|b_i|^q$ are proportional.

Worked Examples

Example 1

For positive reals $a,b,c$ with $a+b+c = 1$, find the minimum of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.

1
Use the $p=q=2$ case (Cauchy–Schwarz, a special case of Hölder) in Engel/sum form, or apply Hölder directly: $\left(\sum \frac{1}{a}\right)\left(\sum a\right) \ge \left(\sum 1\right)^2$ by Cauchy–Schwarz.
2
Substitute $\sum a = 1$ and $\sum 1 = 3$ (three terms): $\left(\frac1a+\frac1b+\frac1c\right)\cdot 1 \ge 3^2 = 9$.
3
Hence $\frac1a + \frac1b + \frac1c \ge 9$.
4
Equality requires $a=b=c$; with $a+b+c=1$ this gives $a=b=c=\frac13$, where $\frac1a+\frac1b+\frac1c = 9$.

Answer:

Minimum value $9$, attained at $a=b=c=\frac{1}{3}$.

Contest level

For positive reals $a, b, c$, prove $(a^3 + b^3 + c^3)(1+1+1)(1+1+1) \ge (a + b + c)^3$, i.e. $9(a^3+b^3+c^3) \ge (a+b+c)^3$.

1
Apply Hölder with three sequences and equal exponents $3,3,3$ (since $\tfrac13+\tfrac13+\tfrac13 = 1$): $\sum a\cdot 1\cdot 1 \le \left(\sum a^3\right)^{1/3}\left(\sum 1^3\right)^{1/3}\left(\sum 1^3\right)^{1/3}$.
2
The right side is $\left(\sum a^3\right)^{1/3}\cdot 3^{1/3}\cdot 3^{1/3}$ and the left side is $a+b+c$.
3
Cube both sides: $(a+b+c)^3 \le \left(\sum a^3\right)\cdot 3\cdot 3 = 9(a^3+b^3+c^3)$.
4
Equality holds when $a=b=c$. (This is exactly the power-mean inequality $M_1 \le M_3$.)

Answer:

$9(a^3+b^3+c^3) \ge (a+b+c)^3$, equality iff $a=b=c$.

Olympiad / Challenge

For positive reals $a, b, c$, prove $\dfrac{a}{\sqrt{a^2+8bc}} + \dfrac{b}{\sqrt{b^2+8ca}} + \dfrac{c}{\sqrt{c^2+8ab}} \ge 1$.

1
Apply the three-factor Hölder with equal weights $\tfrac13,\tfrac13,\tfrac13$ (the $(3,3,3)$ form, since $\tfrac13+\tfrac13+\tfrac13 = 1$) to the three sums $\sum \tfrac{a}{\sqrt{a^2+8bc}}$, $\sum \tfrac{a}{\sqrt{a^2+8bc}}$, and $\sum a(a^2+8bc)$: $\Bigl(\sum \tfrac{a}{\sqrt{a^2+8bc}}\Bigr)\Bigl(\sum \tfrac{a}{\sqrt{a^2+8bc}}\Bigr)\Bigl(\sum a(a^2+8bc)\Bigr) \ge \Bigl(\sum \sqrt[3]{\tfrac{a}{\sqrt{a^2+8bc}}\cdot\tfrac{a}{\sqrt{a^2+8bc}}\cdot a(a^2+8bc)}\Bigr)^3$. Inside the cube root the radicals cancel, $\tfrac{a^2}{a^2+8bc}\cdot a(a^2+8bc) = a^3$, so each summand is $\sqrt[3]{a^3} = a$. Writing $S = \sum \tfrac{a}{\sqrt{a^2+8bc}}$ this is $S^2\cdot\bigl[\sum a(a^2+8bc)\bigr] \ge (a+b+c)^3$.
2
So it suffices to show $(a+b+c)^3 \ge \sum a(a^2+8bc) = a^3+b^3+c^3 + 24abc$.
3
Expand $(a+b+c)^3 = a^3+b^3+c^3 + 3(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b) + 6abc$. The needed inequality reduces to $3(a^2b+a^2c+\cdots+c^2b) + 6abc \ge 24abc$, i.e. $a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \ge 6abc$.
4
That last step is AM–GM on the six terms: their geometric mean is $\sqrt[6]{a^6b^6c^6} = abc$, so the sum is $\ge 6abc$. Hence $S^2 \ge \dfrac{(a+b+c)^3}{\sum a(a^2+8bc)} \ge 1$, giving $S \ge 1$. Equality at $a=b=c$.

Answer:

$\displaystyle\sum \frac{a}{\sqrt{a^2+8bc}} \ge 1$, equality iff $a=b=c$.

Going Deeper

Hölder generalizes to any number of sequences with exponents summing reciprocally to $1$ ($\sum \tfrac{1}{p_k}=1$); the $L^p$-norm integral form $\int |fg| \le \|f\|_p\|g\|_q$ is the analyst's version, and $p=q=2$ is Cauchy–Schwarz.
The 'Hölder for fractions' template — $\bigl(\sum \tfrac{x_i}{y_i}\bigr)\cdot(\cdots) \ge (\sum \text{something})^k$ — is the standard finishing move on hard IMO/shortlist inequalities like the $\sqrt{a^2+8bc}$ problem above.
Pitfall: the exponents must be conjugate ($\tfrac1p+\tfrac1q=1$, both $>1$) and you must keep $a_i, b_i \ge 0$ (or take absolute values). Using $p=q=2$ when the natural pairing is $p=3,\,q=\tfrac32$ gives a weaker, often useless bound.

Spot the Signal

Look for problems where the key step is generalizing Cauchy-Schwarz to bound products and sums in harder inequality problems.
You can describe the hard part as generalizing Cauchy-Schwarz to bound products and sums in harder inequality problems, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the olympiad structure that calls for holder's inequality, then rewrite the givens around it.
Name the relation that makes Holder's Inequality legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Holder's Inequality just because the surface looks familiar; verify the required condition first.
Applying Holder's Inequality because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let holder's inequality reveal the structural theorem; then compute only what remains.

Try it on:

Practice a contest problem where the key step is generalizing Cauchy-Schwarz to bound products and sums in harder inequality problems.