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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 months 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!

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

- 3 months, 3 weeks ago

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.

- 3 months, 2 weeks ago

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

- 3 months, 2 weeks ago

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

- 3 months, 2 weeks ago

I have done it.

- 3 months, 2 weeks ago

Inequality problem 7:

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

- 3 months, 2 weeks ago

Easy it is.

- 3 months, 2 weeks ago

Inequality problem 6:

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

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

- 3 months, 2 weeks ago

Proved it with Jensen

- 3 months, 2 weeks ago

- 3 months, 2 weeks ago

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

- 3 months, 2 weeks ago

OK. :)

- 3 months, 2 weeks ago

If $$x+y+z=1$$ , Prove that

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

- 3 months, 2 weeks ago

Try $$x=y=0$$, $$z=1$$.

- 3 months, 2 weeks ago

But why should i put that value?

- 3 months, 2 weeks ago

- 3 months, 2 weeks ago

If a+b+c=1 , Prove that

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

- 3 months, 2 weeks ago

Try $$a=b=0$$, $$c=1$$.

- 3 months, 2 weeks ago

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

- 3 months, 2 weeks ago

- 3 months, 2 weeks ago

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

- 3 months, 2 weeks ago

Yes which one?

- 3 months, 2 weeks ago

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

- 3 months, 3 weeks ago

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.

- 3 months, 2 weeks ago

- 3 months, 2 weeks ago

Comment deleted 3 months ago

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.

- 3 months, 3 weeks ago

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

- 3 months, 3 weeks ago

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

- 3 months, 2 weeks ago

Can u please write the complete solution

- 3 months, 2 weeks ago

Md zuhair can u please write the complete solution.

- 3 months, 2 weeks ago

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

- 3 months, 2 weeks ago

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

- 3 months, 2 weeks ago

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 . :)

- 3 months, 2 weeks ago

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 $$x=y=z$$.

- 3 months, 2 weeks ago

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

And thanks for the solution

- 3 months, 2 weeks ago

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

- 3 months, 2 weeks ago

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

We have a brilliant group there.

Thank you

- 3 months, 2 weeks ago

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 $$3 - 2 \times (1+3) = 2 \times 4 = -5$$.)

- 3 months, 2 weeks ago

Is my solution correct?

- 3 months, 2 weeks ago

Pardon bro ,,

- 3 months, 2 weeks ago

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

- 3 months, 2 weeks ago

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

- 3 months, 2 weeks ago

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

- 3 months, 2 weeks ago

Okay, Let me see :)

- 3 months, 2 weeks ago