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Note by Dragan Marković
8 months ago

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Hint: Prove that \( x^4 + y^4 \leq x^3 + y^3 \leq x^2 + y^2 \leq x^1 + y^1 \leq x^0 + y^0 = 2 \). Calvin Lin Staff · 7 months, 4 weeks ago

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@Svatejas Shivakumar tbh i was given this problem but couldnt ind any solution well i guess just wrong problem thanks anyway Dragan Marković · 7 months, 4 weeks ago

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@Svatejas Shivakumar can this be proven \(x^{3}+y^{3}\leq 2\) for \(x^{3}+y^{3}\geq x^{3}+y^{4}\) Dragan Marković · 7 months, 4 weeks ago

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@Dragan Marković 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 \(y=0.5\). Svatejas Shivakumar · 7 months, 4 weeks ago

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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, \(x^3+y^3 \leq x+y \leq 2\). Svatejas Shivakumar · 7 months, 4 weeks ago

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@Svatejas Shivakumar 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. Calvin Lin Staff · 7 months, 4 weeks ago

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@Calvin Lin I accidentally posted another incomplete solution but I am not able to edit it. I am also not able to delete this solution. Svatejas Shivakumar · 7 months, 3 weeks ago

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@Svatejas Shivakumar I've deleted your recent comment.

We are working on fixing the "unable to edit / delete comment" bug. Calvin Lin Staff · 7 months, 3 weeks ago

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@Calvin Lin 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 \(x+y \leq 2\).

The answer follows.

In fact, the second inequality is not required. Svatejas Shivakumar · 7 months, 3 weeks ago

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@Svatejas Shivakumar 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, \( x = 1.01, y = 0.9 \).
Calvin Lin Staff · 7 months, 3 weeks ago

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@Calvin Lin 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. Svatejas Shivakumar · 7 months, 3 weeks ago

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@Svatejas Shivakumar 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. Calvin Lin Staff · 7 months, 3 weeks ago

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@Svatejas Shivakumar thank you very much what about the \(x^{3}+y^{3}\geq x^{3}+y^{4}\) Dragan Marković · 7 months, 4 weeks ago

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