Show that $a^4+b^4\geq \frac18$ given that $a+b=1$

Question

Show that $a^4+b^4\geq \frac18$ given that $a+b=1$

$$b=1-a\Rightarrow a^4+b^4=2a^4-4a^3+6a^2-4a+1$$ If we try to find the minimum of a one-variable function,we must solve a 3rd degree equation,on the other hand making perfect squares seems somewhat difficult! Please help.

score 2 · Accepted Answer · answered Aug 21 '17 at 04:43

2

Hint: by the generalized mean inequality: $\;\displaystyle \sqrt[4]{\frac{a^4+b^4}{2}} \ge \frac{a+b}{2}\,$.

answered Aug 21 '17 at 04:43

dxiv

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Show that $a^4+b^4\geq \frac18$ given that $a+b=1$

1 Answers1