# How to deduce the Cauchy–Schwarz inequality from this inequality?

$$\forall(a_1,...,a_n),(b_1,...,b_n)\in\mathbb R^n$$, How to deduce the Cauchy–Schwarz inequality from this inequality : $$\sum_{k=1}^{n}|a_kb_k|\le\frac{1}{2}\left(\sum_{k=1}^{n}a_k^2+\sum_{k=1}^{n}b_k^2\right)$$

I recall that the Cauchy–Schwarz inequality is :$$\left(\sum_{k=1}^{n}a_kb_k\right)^2\le\left(\sum_{k=1}^{n}a_k^2\right)\left(\sum_{k=1}^{n}b_k^2\right)$$

• Apply the given inequality to $(c_k)$ and $(d_k)$ where $c_k=a_k /\sqrt {\sum |a_i|^{2}}$ and $d_k=b_k /\sqrt {\sum |b_i|^{2}}$ Oct 23 '20 at 12:07
• Oh thanks now i understand it ! Oct 23 '20 at 13:00

## 2 Answers

Thanks to Kavi Rama Murthy, I understand how to deduce the Cauchy–Schwarz inequality !

Let $$c_k = a_k/\sqrt{\sum|a_i|^2}$$ and $$d_k = b_k/\sqrt{\sum|b_i|^2}$$

So we have: $$\sum_{k=1}^{n}|c_kd_k|\le\frac{1}{2}\left(\sum_{k=1}^{n}c_k^2+\sum_{k=1}^{n}d_k^2\right)$$ $$\left|\sum_{k=1}^{n}c_kd_k\right|\le\sum_{k=1}^{n}|c_kd_k|\le\frac{1}{2}\left(\sum_{k=1}^{n}c_k^2+\sum_{k=1}^{n}d_k^2\right)$$ $$\left|\sum_{k=1}^{n}c_kd_k\right|\le\frac{1}{2}\left(\sum_{k=1}^{n}c_k^2+\sum_{k=1}^{n}d_k^2\right)$$ $$\left|\sum_{k=1}^{n}c_kd_k\right|^2\le\left(\frac{1}{2}\left(\sum_{k=1}^{n}c_k^2+\sum_{k=1}^{n}d_k^2\right)\right)^2$$ $$\left(\sum_{k=1}^{n}c_kd_k\right)^2\le\left(\frac{1}{2}\left(\sum_{k=1}^{n}c_k^2+\sum_{k=1}^{n}d_k^2\right)\right)^2$$ $$2\left(\sum_{k=1}^{n}c_kd_k\right)^2\le\frac{1}{2}\left(\sum_{k=1}^{n}c_k^2\right)^2+\frac{1}{2}\left(\sum_{k=1}^{n}d_k^2\right)^2+\left(\sum_{k=1}^{n}c_k^2\right)\left(\sum_{k=1}^{n}d_k^2\right)$$ $$2\left(\sum_{k=1}^{n}\frac{a_kb_k}{\sqrt{\sum|a_i|^2\times\sum|b_i|^2}}\right)^2\le\frac{1}{2}\left(\sum_{k=1}^{n}\frac{a_k^2}{\sum|a_i|^2}\right)^2+\frac{1}{2}\left(\sum_{k=1}^{n}\frac{b_k^2}{\sum|b_i|^2}\right)^2+\left(\sum_{k=1}^{n}\frac{a_k^2}{\sum|a_i|^2}\right)\left(\sum_{k=1}^{n}\frac{b_k^2}{\sum|b_i|^2}\right)$$ $$\frac{2}{\sum|a_i|^2\times\sum|b_i|^2}\left(\sum_{k=1}^{n}a_kb_k\right)^2\le\frac{1}{2(\sum|a_i|^2)^2}\left(\sum_{k=1}^{n}a_k^2\right)^2+\frac{1}{2(\sum|b_i|^2)^2}\left(\sum_{k=1}^{n}b_k^2\right)^2+\frac{1}{\sum|a_i|^2\times\sum|b_i|^2}\left(\sum_{k=1}^{n}a_k^2\right)\left(\sum_{k=1}^{n}b_k^2\right)$$ $$\frac{2}{\sum|a_i|^2\times\sum|b_i|^2}\left(\sum_{k=1}^{n}a_kb_k\right)^2\le2$$ $$\left(\sum_{k=1}^{n}a_kb_k\right)^2\le\sum|a_i|^2\times\sum|b_i|^2$$$$\left(\sum_{k=1}^{n}a_kb_k\right)^2\le\sum a_k^2\times\sum b_k^2$$

$${\bf Hint:}\quad 0 \leq p^2 - 2pq+ q^2$$.