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Let $$x,y$$ be positive real numbers such that $$x^{3}+y^{3}\geq x^{4}+y^{4}$$ Prove that $$x^{3}+y^{3}\leq 2$$.

Note by Dragan Marković
5 months, 3 weeks ago

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Hint: Prove that $$x^4 + y^4 \leq x^3 + y^3 \leq x^2 + y^2 \leq x^1 + y^1 \leq x^0 + y^0 = 2$$. Staff · 5 months, 2 weeks ago

@Svatejas Shivakumar tbh i was given this problem but couldnt ind any solution well i guess just wrong problem thanks anyway · 5 months, 3 weeks ago

@Svatejas Shivakumar can this be proven $$x^{3}+y^{3}\leq 2$$ for $$x^{3}+y^{3}\geq x^{3}+y^{4}$$ · 5 months, 3 weeks ago

I think that you mean $$x^3+y^3 \geq x^4+y^4$$ instead of $$x^3+y^3 \geq x^3+y^4$$ for $$x+y \leq 2$$.

The converse is not true. For a counterexample, take $$x=1.5$$ and $$y=0.5$$. · 5 months, 3 weeks ago

Let's assume WLOG, that $$x \ge y$$.

We have $$x^3(x-1)+y^3(y-1) \leq 0$$. So, $$x \leq 1$$.

By Muirhead's inequality, $$x^4+y^4 \leq x^3+y^3 \leq x^3y+xy^3$$ so $$x^3(y-1)+y^3(x-1) \leq 0$$. So, $$y \leq 1$$.

We know that $$a^3 \leq a$$ for a which are positive real numbers.

So, $$x^3+y^3 \leq x+y \leq 2$$. · 5 months, 3 weeks ago

It is not true that $$x \leq 1$$. As a counter example, $$1.01^3 + 0.9^3 \approx 1.76 > 1.01^4 + 0.9^4 \approx 1.70$$

Do you see the error in your reasoning? Hint: What is the sign of $$y - 1$$?

Note: It is also not true that $$a^ 3 \leq a$$ for positive real numbers, but that wasn't what you intended to say. Staff · 5 months, 2 weeks ago

I accidentally posted another incomplete solution but I am not able to edit it. I am also not able to delete this solution. · 5 months, 2 weeks ago

We are working on fixing the "unable to edit / delete comment" bug. Staff · 5 months, 2 weeks ago

Yes I forgot to take the sign of y-1.

We must have $$x-1 \leq |y-1|$$. Solving we get $$x \leq y$$(false due to our assumption) or $$x+y \leq 2$$.

In fact, the second inequality is not required. · 5 months, 2 weeks ago

There are now more issues.

1. Why must we have $$x - 1 \leq |y - 1 |$$? That claim has not been justified as yet.
2. Even if that is true, it does not lead to $$x \leq y$$. For example, $$x = 1.01, y = 0.9$$.
Staff · 5 months, 2 weeks ago

I meant that $$x \leq y$$ if $$y \geq 1$$ and $$x+y \leq 2$$ if $$y \leq 1$$. The reason the inequality holds is because x^3>y^3 due to our assumption. · 5 months, 2 weeks ago

What you meant, is very different from what (you think) you wrote.

Given that your solution has numerous issues, I advise you to rewrite it from the start so that it can be clearly understood, instead of having others guess at what the missing steps are. Staff · 5 months, 2 weeks ago

thank you very much what about the $$x^{3}+y^{3}\geq x^{3}+y^{4}$$ · 5 months, 3 weeks ago