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

**Topics:**

- Inequality

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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

Prove or disprove!

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

Log in to reply

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.

Log in to reply

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

Log in to reply

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

Log in to reply

I have done it.

Log in to reply

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

Log in to reply

Easy it is.

Log in to reply

Inequality problem 6:Given positive reals \(a,b,c\) satisfying \(a,b,c\le 1\) and \(a+b+c\ge 1\).

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

Log in to reply

Proved it with Jensen

Log in to reply

Can you show your work?

Log in to reply

Log in to reply

Log in to reply

Please Help

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

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

Log in to reply

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

Log in to reply

But why should i put that value?

Log in to reply

Log in to reply

If a+b+c=1 , Prove that

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

Log in to reply

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

Log in to reply

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

Log in to reply

Log in to reply

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

Log in to reply

Yes which one?

Log in to reply

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.

Log in to reply

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

Log in to reply

Comment deleted 2 weeks ago

Log in to reply

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

Log in to reply

Can u please write the complete solution

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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 . :)

Log in to reply

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 \(x=y=z\).

Log in to reply

And thanks for the solution

Log in to reply

Log in to reply

We have a brilliant group there.

Thank you

Log in to reply

Log in to reply

Log in to reply

Log in to reply

@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.

Log in to reply

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

Log in to reply

Okay, Let me see :)

Log in to reply