https://aops.com/community/p11366986 programjames1 wrote: By Cauchy-Schwartz, $$(a^n+1)(b^n+1)\ge (a^\frac{n}2+b^\frac{n}2)^2$$And so it suffices to prove that $a^\frac{n}2+b^\frac{n}2\ge 2$ which is true by the Power-Mean inequality.

Okay, so I got the hint to do Cauchy Schwarz and so we will do it.

jasperE3 wrote: https://aops.com/community/p11366986 programjames1 wrote: By Cauchy-Schwartz, $$(a^n+1)(b^n+1)\ge (a^\frac{n}2+b^\frac{n}2)^2$$And so it suffices to prove that $a^\frac{n}2+b^\frac{n}2\ge 2$ which is true by the Power-Mean inequality. Did you just post the problem and then the solution immediately after? If so what was the necessity to post the question in the first place?

What was the necessity to not post the solution immediately after the question? Think

centslordm wrote: What was the necessity to not post the solution immediately after the question? Think So others could solve without a solution given... That is quite frankly a non-sensical question, and I find it quite post-farmy as it did not answer my question and was completely unrelated to the topic. What is the intent of the post? Was it to look for other solutions? I'm very confused... Did @below just post a completely new question that had nothing to do with the topic? Shouldn't it be a new post?

Prove that for all $a,b>0$ with $a+b=2$, we have $$\left(a+1\right)\left(b^2+1\right) \ge\frac{4}{9}(9-\sqrt 6)$$Equality holds when $a=1+\sqrt{\frac{2}{3}},b=1-\sqrt{\frac{2}{3}}.$ Prove that for all $a,b>0$ with $a+b^2=2$, we have $$\left(a+1\right)\left(b+1\right) \le\frac{4}{27}(13+5\sqrt{10})$$Equality holds when $a=\frac{7+2\sqrt{10}}{9},b=\frac{\sqrt{10}-1}{3}.$ Prove that for all $a,b>0$ with $a+2b=2$, we have $$\left(a^2+1\right)\left(b^2+1\right) >\frac{9}{5}$$

jasperE3 wrote: Prove that for all $a,b>0$ with $a+b=2$, we have $$\left(a^n+1\right)\left(b^n+1\right)\ge4$$for all $n\in\mathbb N_{\ge2}$. The solution below is incorrect. I am not sure how to fix it. If you have any suggestions, you can pm me or reply.

bigbrain123 wrote: Expanding, we have $a^n b^n +a^n+b^n \geq 3$. From AM-GM, we have $ab \leq (\frac{a+b}{2})^2 = 1$. Thus, it suffices to prove $a^n + b^n \geq 2$. Um, no? The inequality $ab\le 1$ is correct, but points in the wrong direction for your estimate...

Sorry @above, what do you mean? What is pointing in the wrong direction? I replaced $ab$ with its maximum $1$ on the lesser side, so if the maximum of the LHS is lesser than the RHS, then we proved it... right?

Unfortunately, it is not the "lesser side"...

:faceslap:. Ah yes, what a big silly. I suppose that I'll try again. Thanks for your correction!

bigbrain123 wrote: I'm not too sure if my contradiction argument is rigorous Unfortunately, reversing all the logical arrows by formulating an argument as a proof by contradiction does not turn a wrong argument in a correct one. From $a^nb^n+a^n+b^n<3$ and $ab \le 1$ you simply cannot deduce that $a^n+b^n<2$. (There is also a minor error in your application of the mean inequality where the denominator should be $2$ and not $n$, but that one is harmless.)

Any solution...?

lazizbek42 wrote: Any solution...? See #2,#3.

Jasper E3 I think wrong solution.

By Cauchy-Schwartz, $$(a^n+1)(b^n+1)\ge (a^\frac{n}2+b^\frac{n}2)^2$$And so it suffices to prove that $a^\frac{n}2+b^\frac{n}2\ge 2$ But $2* (a^\frac{n}2+b^\frac{n}2) +2*n - 4$ $=$ $(2* a^\frac{n}2 + n - 2) +(2* b^\frac{n}2 + n - 2)\ge n*a + n*b = 2n$ $by$ $AM-GM$ So $a^\frac{n}2+b^\frac{n}2\ge 2$ $ and$ $ we$ $ are $ $ done.$

lazizbek42 wrote: Jasper E3 I think wrong solution. #3 is wrong but #2 is fine.

What's wrong with #3?

This is a reversed inequality: Quote: we have $a^{\frac{n}{2}}+b^{\frac{n}{2}} \geq 2\sqrt{(ab)^{\frac{n}{2}}}$. The maximum value of $ab$ is $1$, so the maximum value of this is just $2\sqrt{1}=2$ as desired.

All solutions wrong or not complete

The solution in post 2 is correct.

exactly this error, Solution 2.

$(a^n+1)(1+b^n)(1+1)(1+1)...(1+1)\ge (a+b)^n=2^n$ $(a^n+1)(1+b^n)\ge 4$ Holder's inequality