Prove $4a^2+b^2+1\ge2ab+2a+b$
$4a^2+b^2+1-2ab-2a-b\ge0$
$(2)^2(a)^2+(b)^2+1-2ab-2a-b\ge0$
Any help from here? I am not seeing how this can be factored
Asked
Active
Viewed 83 times
5
Michelle Stasso
- 193
-
1Complete the square in $a$ first and then in $b$. – A.Γ. Nov 29 '16 at 12:49
4 Answers
5
As a function of $a$, the minimum of the quadratic $4a^2+b^2+1-(2ab+2a+b)$ is $\frac34 (b - 1)^2 \ge 0$.
More precisely, $$4a^2+b^2+1-(2ab+2a+b)=\left(2a-\frac{b+1}{2}\right)^2+\frac34 (b - 1)^2 \ge 0$$
lhf
- 216,483
-
2To add some explanation: Consider the function $f(a)=4a^2+b^2+1−2ab−2a−b$. Since it is a polynomial of degree 2 with positive leading coefficient it has its minimum at the root of $f'(a)$, which is $a=\frac{1}{4}(b+1)$. – ctst Nov 29 '16 at 12:49
5
Let $x=(2a,b,1)$ and $y=(1,2a,b)$. Then by, Cauchy-Schwarz,
$2ab+2a+b= <x,y> \le ||x||*||y||=\sqrt{4a^2+b^2+1}*\sqrt{4a^2+b^2+1}=4a^2+b^2+1$
Fred
- 77,394
3
Multiply by $2$, then we get: $$8a^2+2b^2+2 \ge 4ab+4a+2b$$ and rearranging we get: $$(4a^2-4ab+b^2)+(b^2-2b+1)+(4a^2-4a+1) \ge 0$$ and then $$(2a-b)^2+(b-1)^2+(2a-1)^2 \ge 0$$
Arnaldo
- 21,342
2
High school solution: $$ 4a^2+b^2+1-2ab-2a-b=\frac{(2a-1)^2+(2a-b)^2+(b-1)^2}{2}\ge 0. $$
A.Γ.
- 29,518