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

**Topics:**

- Inequality

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## Comments

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TopNewestInequality problem 3:If \(a, b, c>0,\) such that \(ab+bc+ca=1\). Where \(a,b,c\) are distinct.

Prove or disprove!

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

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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.

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

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

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I have done it.

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Inequality problem 7:Suppose \(x,y,z \ge 0.\) Prove that \[\sqrt{x^2+\frac{1}{y^2}}+\sqrt{y^2+\frac{1}{z^2}}+\sqrt{z^2+\frac{1}{x^2}}\ge 3\sqrt{2}\]

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Easy it is.

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

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Proved it with Jensen

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Can you show your work?

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Please Help

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

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

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

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But why should i put that value?

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If a+b+c=1 , Prove that

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

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

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

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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

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Yes which one?

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

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.

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

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Comment deleted 3 months 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 Disproveproblem in which you just need to give an example.Log in to reply

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

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

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Can u please write the complete solution

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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 . :)

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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 \(x=y=z\).

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And thanks for the solution

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We have a brilliant group there.

Thank you

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@Sharky Kesa , what do you think?

I think it is Ok.Log in to reply

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

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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

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Okay, Let me see :)

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