1

Prove that if $a$ and $b$ satisfy $2a+4b=1$, then $a^2+b^2\ge \frac {1}{20}$

The only idea I have is that I could apply Cauchy-Schwartz but i don't see how, any hints?

Michael Rozenberg
  • 194,933
Trobeli
  • 3,242
  • 5
    Apply it to the vectors $(a,b)$ and $(2,4)$. You get that $(a^2+b^2)(2^2+4^2)\geq(2a+4b)=1$ – orole Feb 06 '18 at 19:09
  • Or by brute force, take $2a=1-4b$ from the first equation and substitute into the second one, then work out the quadratic $,5\left((1-4b)^2+4b^2\right) - 1 = 4,(5b-1)^2,$. – dxiv Feb 06 '18 at 19:11

6 Answers6

5

We need to prove that $$a^2+b^2\geq\frac{(2a+4b)^2}{20}$$ or $$(2a-b)^2\geq0.$$ Also, we can use C-S: $$a^2+b^2=\frac{1}{5}(1^2+2^2)(a^2+b^2)\geq\frac{(a+2b)^2}{5}=\frac{1}{20}.$$

Michael Rozenberg
  • 194,933
2

An alternative approach is to use AM-QM inequality as $$\left(\frac{2a + b + b + b +b }{5}\right)^2 \le \frac{4a^2+b^2+b^2+b^2+b^2}{5} \implies a^2+b^2 \ge \frac{1}{20}.$$

Math Lover
  • 15,153
2

Using coordinate geometry

The perpendicular distance of $2x+4y=1$ from the origin is given by

$$d=\frac{1}{\sqrt{2^2+4^2}}=\frac{1}{\sqrt{20}}$$

By $\sqrt{x^2+y^2} \ge d$ gives

$$x^2+y^2 \ge \frac{1}{20}$$

Ng Chung Tak
  • 18,990
2

Since $$ a=\frac12-2b $$ we have $$ \begin{align} a^2+b^2 &=\frac14-2b+5b^2\\ &=5\left(b-\frac15\right)^2+\frac1{20} \end{align} $$

robjohn
  • 345,667
1

The minimal value of the parabola $$ a^2+b^2=\left( \frac 1 2-2\,b \right) ^{2}+{b}^{2}=\frac 1 4-2\,b+5\,{b}^{2} $$ is at its vertex $b=\dfrac{1}{5}$ and it equals exactly $\dfrac{1}{20}.$

Leox
  • 8,120
0

this is equivalent to $$(10a-1)^2\geq 0$$ by inserting $$b=\frac{1}{4}-\frac{1}{2}a$$ in the left Hand side and factorizing! and the term above is $$\frac{5}{4}a^2+\frac{1}{16}-\frac{1}{4}a-\frac{1}{20}$$

Dr. Sonnhard Graubner
  • 95,283