Question

if a,b are positive numbers and a+b=1, the minimum value of a^4+b^4=? I will surely mark as brainliest do not scam

Answer 1

Answer:

Step-by-step explanation:

$(a_{1} b_{1} + a_{2} b_{2} + . . . + a_{n} b_{n})^{2} \leq (a^{2} _{1} + a^{2} _{2} + . . . + a^{2} _{n}) (b^{2} _{1} + b^{2} _{2} +b^{2} _{3} . . . + b^{2} _{n})$

So

$(a + b)^{2} \leq (a^{2} + b^{2} ) (1 + 1)$

$1 \leq 2(a^{2} + b^{2} )$

$(a^{2} + b^{2} ) \geq \frac{1}{2}$

Use the inequality… again:

$(a^{2} + b^{2} )^{2} \leq (a^{4} + b^{4} ) (1 + 1)$

$\frac{1}{4} \leq 2(a^{4} + b^{4} )$

$a^{4} + b^{4} \leq \frac{1}{8}$

So the small possible value of $a^{4} + b^{2} is \frac{1}{8}$

Oh, I (almost) forgot. The inequality become an inequality if $\frac{a_{1} }{b_{1} } = \frac{a_{2} }{b_{2} } = . . . = \frac{a_{n} }{b_{n} }$

So  $a^{4} + b^{2} = \frac{1}{8}$  when $\frac{a}{1 } = \frac{b}{1} or a = b$

But  $a + b = 1$  so $a = b = \frac{a + b}{2} = \frac{1}{2}$

Hope this helps you !!!