Deduce the Cauchy–Schwarz inequality from this inequality

Question

$\forall(a_1,…,a_n),(b_1,…,b_n)\in\mathbb R^n$, 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 :$$\left(\sum_{k=1}^{n}a_kb_k\right)^2\le\left(\sum_{k=1}^{n}a_k^2\right)\left(\sum_{k=1}^{n}b_k^2\right)$$

Answer 1

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