Why is the Cauchy Schwarz inequality a special case of Holder's inequality?

Cauchy Schwarz inequality is said to be a special case of Holder's inequality when $p=2,q=2$. $(1)$ is the Holder's inequality and $(2)$ is the Cauchy Schwarz inequality. I cannot see how putting $p=2$ and $q=2$ in $(1)$ yields $(2)$. i.e. saying $|a+b|≤c$ isn't the same as saying $|a|+|b|≤c$.

• I don't follow what the question is exactly. Can you explain a bit better? – Cameron Williams Feb 14 '18 at 16:45
• You have a low resolution image of the inequalities that you are interested in, one of which is labeled "Hölder's inequality," and the other is labeled "Hölder's sum inequality." The second looks like Cauchy-Schwarz to me, which looks like a direct application of the first inequality (after squaring both sides). I don't understand your question... – Xander Henderson Feb 14 '18 at 16:46
• @XanderHenderson no it isn't. It says a special case of Holder's sum inequality. Which is Cauchy Schwarz inequality. – Hrit Roy Feb 14 '18 at 16:55
• @CameronWilliams I edited it. Could you check it once more? – Hrit Roy Feb 14 '18 at 17:00
• You are not a new user. Why are you posting lousy images instead of typing the equations in LaTeX? – user21820 May 10 '18 at 8:09

Pugging $p=q=2$ into Holder's inequality you obtain $$\sum_{k=1}^n\lvert a_kb_k\rvert \leq \left(\sum_{k=1}^n\lvert a_k\rvert^2\right)^{1/2}\left(\sum_{k=1}^n\lvert b_k\rvert^2\right)^{1/2}$$ Squaring either side you obtain exactly Cauchy's inequality; $$\left(\sum_{k=1}^n\lvert a_kb_k\rvert\right)^2 \leq \left(\sum_{k=1}^n\lvert a_k\rvert^2\right)\left(\sum_{k=1}^n\lvert b_k\rvert^2\right)$$
• @HritRoy Sure it does. However, if the $a_k$ and $b_k$ are real, then $|a_k|^2 = a_k^2$, so they are often elided in real spaces. – Xander Henderson Feb 14 '18 at 17:01