Verified answer

[tex]\displaystyle \sf \frac{1}{2}\cdot \sum^n_{k=1} \left(a_k^2+b_k^2\right) \geq \sum^n_{k=1}a_kb_k \\\\\\\ k = 1 : \\\\ \to a_1^2+b_1^2\geq 2a_1b_1 \to a_1^2-2a_1b_1+b_1^2\geq 0 \to \underbrace{\sf (a_1-b_1)^2}_{\geq 0\ \forall\ a_1,b_1 \in\mathbb{R} }\geq 0 \\\\\\[/tex]

[tex]\displaystyle \sf k = 2 : \\\\ (a_1-b_1)^2+a_2^2+b_2^2\geq 2a_2b_2 \\\\ (a_1-b_1)^2+a_2-2a_2b_2+b^2_2\geq 0 \\\\ \underbrace{\sf (a_1-b_1)^2}_{\geq 0\ \forall\ a_2,b_2 \in\mathbb{R} } + \underbrace{\sf (a_2-b_2)^2}_{\geq 0\ \forall\ a_2,b_2 \in\mathbb{R} } \geq 0 \ \checkmark[/tex]

.....

[tex]\displaystyle \sf k = n : \\\\ \underbrace{\sf \underbrace{\sf (a_1-b_1)^2+(a_2-b_2)^2+(a_3-b_3)^2+...}_{\displaystyle \geq 0 \ \forall \ a_1,b_1,...,a_{n-1},b_{n-1}\ \in\ \mathbb{R}}+\underbrace{\sf (a_n-b_n)^2}_{\geq 0 \ \forall \ a_n, b_n\ \in\ \mathbb{R}}}_{\displaystyle \geq 0 } \geq 0[/tex]
Então :
[tex]\Large\boxed{\ \sf \displaystyle \ \frac{1}{2}\left[\cdot \sum^n_{k=1} a_k^2+\sum^n_{k=1} b_k^2\right] \geq \sum^n_{k=1}a_kb_k\ } \ \checkmark[/tex]

Sejam x e y reais, façamos :

[tex]\displaystyle \sf (x-y)^2\geq 0 \\\\\ x^2+y^2-2xy\geq 0 \\\\ x^2+y^2 \geq 2xy \\\\ \Large\boxed{\sf \ xy \leq \frac{(x^2+y^2)}{2} \ }\checkmark[/tex]