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

@Svatejas Shivakumar
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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
·
9 months, 3 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.
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Svatejas Shivakumar
·
9 months, 3 weeks ago

@Calvin Lin
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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.
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Svatejas Shivakumar
·
9 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
·
9 months, 3 weeks ago

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TopNewestHint: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 · 9 months, 3 weeks agoLog in to reply

@Svatejas Shivakumar tbh i was given this problem but couldnt ind any solution well i guess just wrong problem thanks anyway – Dragan Marković · 9 months, 3 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 · 9 months, 3 weeks ago

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The converse is not true. For a counterexample, take \(x=1.5\) and \(y=0.5\). – Svatejas Shivakumar · 9 months, 3 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 · 9 months, 3 weeks ago

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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 · 9 months, 3 weeks ago

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– Svatejas Shivakumar · 9 months, 3 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.Log in to reply

We are working on fixing the "unable to edit / delete comment" bug. – Calvin Lin Staff · 9 months, 3 weeks ago

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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 · 9 months, 3 weeks ago

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– Svatejas Shivakumar · 9 months, 3 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.Log in to reply

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 · 9 months, 3 weeks ago

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– Dragan Marković · 9 months, 3 weeks ago

thank you very much what about the \(x^{3}+y^{3}\geq x^{3}+y^{4}\)Log in to reply