Help in proving an inequality

Question

Show that $$a^4 + b^4\ge\frac{1}{8}$$ if $a+b=1.$

Answer 1

Hint: Note that $ 2(a^2 +b^2) \geq (a+b)^2, ~\forall a,b\in\mathbb{R}$ and use it twice to derive the desired inequality.

Answer 2

hint: use CS inequality twice: $(ab + cd)^2 \leq (a^2+c^2)(b^2+d^2)$

Answer 3

$$\sqrt{\frac{x^2+y^2}{2}}\ge\frac{x+y}2 \Rightarrow x^2+y^2\ge\frac12(x+y)^2$$ Then $$a^4+b^4\ge\frac12(a^2+b^2)^2\ge\frac12\cdot\left(\frac12(a+b)^2\right)^2=\frac18$$

Answer 4

$\text {let: } k = \frac {(b-a)}{2}\\ \text {then: } b = \frac {(b+a)}{2} + \frac {(b-a)}{2} = \frac 12 + k\\ a = \frac {(b+a)}{2} - \frac {(b-a)}{2} = \frac 12 - k$

$a^4 + b^4\\ (\frac 12 - k)^4 + (\frac 12 + k)^2\\ [(\frac 12)^4 - 4(\frac 12)^3k + 6(\frac 12)^2k^2 -4(\frac 12)k^3 + k^4] + [(\frac 12)^4 + 4(\frac 12)^3k + 6(\frac 12)^2k^2 +4(\frac 12)k^3 + k^4]\\ (\frac 12)^4 + 6(\frac 12)^2k^2 + k^4\\ \frac 18 + 6(\frac 12)^2k^2 + k^4$

$k^2>0, k^4 >0$ for all k.

$\frac 18 + 6(\frac 12)^2k^2 + k^4 > \frac 18$