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Inequality Problem and solution

Give some proof problems below(So others can give solutions). Which could help to increase our problem-solving skills.

Topics:

  • Inequality

Note by Munem Sahariar
4 weeks ago

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Inequality problem 3:

If \(a, b, c>0,\) such that \(ab+bc+ca=1\). Where \(a,b,c\) are distinct.

Prove or disprove!

\[\frac{a^3+b^3}{(a-b)^2}+\frac{b^3+c^3}{(b-c)^2}+\frac{c^3+a^3}{(c-a)^2}< 5\]

Munem Sahariar - 3 weeks ago

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I mean, this inequality is obviously false. Just bring \(a, b, c\) close to each other (I did \(a=0.57, b=0.58, c\approx 0.58207 \neq b\) such that \(ab+bc+ca=1\) (checked by WA). Then the value of the expression is \(96501.7...\). So now it suffices to show the opposite is true. Will post work later, but I need to sleep.

Sharky Kesa - 2 weeks, 2 days ago

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@Sharky Kesa I know it's false. You just have to disprove it. \(\ddot \smile\)

Munem Sahariar - 2 weeks, 1 day ago

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You need to mention here that a,b,c are distinct

Md Zuhair - 2 weeks, 2 days ago

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I have done it.

Munem Sahariar - 2 weeks, 2 days ago

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Inequality problem 7:

Suppose \(x,y,z \ge 0.\) Prove that \[\sqrt{x^2+\frac{1}{y^2}}+\sqrt{y^2+\frac{1}{z^2}}+\sqrt{z^2+\frac{1}{x^2}}\ge 3\sqrt{2}\]

Munem Sahariar - 1 week, 5 days ago

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Easy it is.

Md Zuhair - 1 week, 5 days ago

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Inequality problem 6:

Given positive reals \(a,b,c\) satisfying \(a,b,c\le 1\) and \(a+b+c\ge 1\).

Prove that \[\frac{a}{3-2a}+\frac{b}{3-2b}+\frac{c}{3-2c}\ge\frac37.\]

Munem Sahariar - 2 weeks, 1 day ago

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Proved it with Jensen

Md Zuhair - 1 week, 5 days ago

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Can you show your work?

Munem Sahariar - 1 week, 5 days ago

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@Munem Sahariar If u dont mind can I show it at evening. I am using mobile and its difficult to write latex in mobile

Md Zuhair - 1 week, 5 days ago

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@Md Zuhair OK. :)

Munem Sahariar - 1 week, 5 days ago

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Please Help

If \(x+y+z=1\) , Prove that

\[\displaystyle{\sum_{cyc}^{x,y,z} \dfrac{x^2+yz}{2x^3+7yz} \geq \dfrac{18}{23}}\]

Md Zuhair - 2 weeks, 2 days ago

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Try \(x=y=0\), \(z=1\).

Sharky Kesa - 2 weeks, 2 days ago

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But why should i put that value?

Md Zuhair - 2 weeks, 2 days ago

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@Md Zuhair Those values cause a contradiction.

Sharky Kesa - 2 weeks, 2 days ago

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If a+b+c=1 , Prove that

\[\displaystyle{\sum_{cyc}^{x,y,z} \dfrac{a^3+bc}{a^2+bc} \geq 2}\]

Md Zuhair - 2 weeks, 2 days ago

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Try \(a=b=0\), \(c=1\).

Sharky Kesa - 2 weeks, 2 days ago

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Why should I put values? It doesnt show if it is minimum or max

Md Zuhair - 2 weeks, 2 days ago

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@Md Zuhair Those values cause a contradiction.

Sharky Kesa - 2 weeks, 2 days ago

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Hello. I want your contribution here.

@Sharky Kesa @Md Zuhair @Atomsky Jahid @Chew-Seong Cheong @Pi Han Goh @Zach Abueg @Steven Yuan @Mark Hennings @Rahil Sehgal @Brian Charlesworth @Brandon Monsen

Munem Sahariar - 2 weeks, 2 days ago

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Yes which one?

Md Zuhair - 2 weeks, 2 days ago

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Inequality Problem 2: Given that \(a,b,c>0\). Prove that \[\large (a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3)\geq(a+b+c)^3\]

Vilakshan Gupta - 3 weeks ago

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USAMO 2004, if I'm not mistaken. \[(a^5 - a^2 + 3) = (a^3 + 2 + (a^3 - 1)(a^2 - 1)) \geq (a^3+1+1).\] Similarly, \((b^5 - b^2 + 3) \geq (1+b^3 + 1)\) and \((c^5 - c^2 + 3) \geq 1 + 1 + c^3\).

Thus, we have \[(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \geq (a^3 + 1 + 1)(1 + b^3 + 1)(1 + 1 + c^3) \geq (a+b+c)^3\] where the final statement is true by Cauchy-Schwarz/Holder.

Sharky Kesa - 2 weeks, 2 days ago

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Could you please help me by doing the problems posted by me :P

Md Zuhair - 2 weeks, 2 days ago

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Comment deleted 2 weeks ago

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Hey, this is not a proof. You just took one example and you are saying that it is true for all real values of \(x\).How are you sure that there is no other value which will dissatisfy the condition. This is not a Prove or Disprove problem in which you just need to give an example.

Vilakshan Gupta - 3 weeks ago

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Inequality Problem 1: \(x,y,z\) are positive real numbers and \(xyz\geq 1\) , Prove that \[\large \frac{x^5-x^2}{x^5+y^2+z^2} + \frac{y^5-y^2}{y^5+z^2+x^2}+\frac{z^5-z^2}{z^5+x^2+y^2}\geq0\]

Vilakshan Gupta - 3 weeks ago

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DONE THIS ONE :P

Just add y^2+z^2 in the numerator of the 1st, and correspondingly subtract, then you will get(after a little more calcs)

\(\displaystyle{3-\sum_{cyc}^{x,y,z} \dfrac{(x^2+y^2+z^2)}{x^5+y^2+z^2}}\)

Now take \((x^2+y^2+z^2)\) common and then use Titu's Lemma like

\(\displaystyle{3-(x^2+y^2+z^2) * \sum_{cyc}^{x,y,z} \dfrac{1}{x^5+y^2+z^2} \geq \dfrac{3^2}{x^5+y^5+z^5+2(x^2+y^2+z^2)}}\)

Now do it on your own

Md Zuhair - 2 weeks, 2 days ago

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Can u please write the complete solution

Vilakshan Gupta - 2 weeks, 2 days ago

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@Vilakshan Gupta Md zuhair can u please write the complete solution.

Vilakshan Gupta - 2 weeks, 2 days ago

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@Vilakshan Gupta Hey!! Yeah, Will write a solution, but do you think I am going on the correct track?

Md Zuhair - 2 weeks, 2 days ago

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@Md Zuhair I don't know,that's why i asked.

Vilakshan Gupta - 2 weeks, 2 days ago

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@Vilakshan Gupta Okay, Let me give the solution

Just add y^2+z^2 in the numerator of the 1st, and correspondingly subtract, then you will get(after a little more calcs)

\(\displaystyle{3-\sum_{cyc}^{x,y,z} \dfrac{(x^2+y^2+z^2)}{x^5+y^2+z^2}}\)

Now take \((x^2+y^2+z^2)\) common and then use Titu's Lemma like

\(\displaystyle{3-(x^2+y^2+z^2) * \sum_{cyc}^{x,y,z} \dfrac{1}{x^5+y^2+z^2} \geq \dfrac{3^3}{x^5+y^5+z^5+2(x^2+y^2+z^2)}}\)

Now given that \(xyz \geq 1\) this simply means

\(x^2+y^2+z^2 \geq 3\)

\(x^5+y^5+z^5 \geq 3\)

\(x^5+y^5+z^5+2(x^2+y^2+z^2) \geq 9\)

\(\dfrac{1}{x^5+y^5+z^5+2(x^2+y^2+z^2)} \leq \dfrac{1}{9}\)

NOW HERE IS WHERE I DOUBT MY SOLUTION...

\(\displaystyle{3-(x^2+y^2+z^2) * \sum_{cyc}^{x,y,z} \dfrac{1}{x^5+y^2+z^2} \geq 3*\dfrac{3^2}{x^5+y^5+z^5+2(x^2+y^2+z^2)} \geq 3-\dfrac{3^3}{9} = 0}\)

BUT I GUESS ITS WRONG . :)

Md Zuhair - 2 weeks, 2 days ago

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@Md Zuhair You had the right starting, but here's a better way to finish:

We have to prove \(\displaystyle \sum_{\text{cyc}} \dfrac{x^5 - x^2}{x^5 + y^2 + z^2} \geq 0\). \(\dfrac{x^5 - x^2}{x^5 + y^2 + z^2} = 1 - \dfrac{x^2+y^2+z^2}{x^5+y^2+z^2}\).

Thus, it suffices to prove \(\displaystyle \sum_{\text{cyc}} \dfrac{x^2+y^2+z^2}{x^5 + y^2 + z^2} \leq 3\). By Cauchy-Schwarz, we have \((x^5 + y^2 + z^2) \left ( \frac{1}{x} + y^2 + z^2 \right ) \geq (x^2 + y^2 + z^2)^2\), so \(\dfrac{\frac{1}{x} + y^2 + z^2}{x^2 + y^2 + z^2} \geq \frac{x^2 + y^2 + z^2}{x^5 + y^2 + z^2}\).

Thus, it suffices to prove \(\displaystyle \sum_{\text{cyc}} \dfrac{\frac{1}{x} + y^2 + z^2}{x^2 + y^2 +z^2} \leq 3 \iff \displaystyle \sum_{\text{cyc}} \dfrac{\frac{1}{x}}{x^2 + y^2 + z^2} \leq 1 \iff \displaystyle \sum_{\text{cyc}} \dfrac{1}{x} \leq x^2 + y^2 + z^2\).

However, by AM-GM, we have \(\displaystyle \sum_{\text{cyc}} \dfrac{1}{x} \leq xy+yz+zx \leq x^2 + y^2 + z^2\), where equality occurs \(x=y=z\).

Sharky Kesa - 2 weeks, 1 day ago

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@Sharky Kesa So is my solution correct or not? I may have done it with different approach. :)

And thanks for the solution

Md Zuhair - 2 weeks, 1 day ago

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@Md Zuhair No, I don't believe it is. That area where you said you doubt is incorrect.

Sharky Kesa - 2 weeks, 1 day ago

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@Sharky Kesa Okay. Thanks. And do you have any kind of social media like Whatsapp or something.

We have a brilliant group there.

Thank you

Md Zuhair - 2 weeks, 1 day ago

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@Md Zuhair Can you parse your statements properly? I think I know what you're trying to say, but I need it verified. (Don't do things like \(3 - 2 \times (1+3) = 2 \times 4 = -5\).)

Sharky Kesa - 2 weeks, 1 day ago

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@Sharky Kesa Is my solution correct?

Md Zuhair - 2 weeks, 1 day ago

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@Sharky Kesa Pardon bro ,,

Md Zuhair - 2 weeks, 1 day ago

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@Md Zuhair I think it is Ok. @Sharky Kesa , what do you think?

Vilakshan Gupta - 2 weeks, 2 days ago

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@Md Zuhair you can try this problem and then write the solution.

Vilakshan Gupta - 2 weeks, 2 days ago

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Its quite simple as it seems, Idk I did it without any paper work, May be i have done wrong calcs :P

Md Zuhair - 2 weeks, 2 days ago

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Okay, Let me see :)

Md Zuhair - 2 weeks, 2 days ago

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