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nguyenhtchuyentb

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nguyenhtchuyentb

#1 h Sep 19, 2010, 2:10 PM • 2 Y

Y by Adventure10, Mango247

Let a,b,c be real posives such that a+b+c=3. Prove that
$\sqrt[3]{a} +\sqrt[3]{b} +\sqrt[3]{c} \geq ab+bc+ca$

This post has been edited 1 time. Last edited by nguyenhtchuyentb, Sep 19, 2010, 2:26 PM

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truongtansang89

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truongtansang89

#2 h Sep 19, 2010, 2:23 PM • 2 Y

Y by Adventure10, Mango247

nguyenhtchuyentb wrote:

Let a,b,c be posives such that a+b+c=3. Prove that
\sqrt[3]{a} +\sqrt[3]{b} +\sqrt[3]{c} \ge ab+bc+ca

Edit : Sorry, I misread.

This post has been edited 1 time. Last edited by truongtansang89, Sep 19, 2010, 3:01 PM

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nguyenhtchuyentb

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nguyenhtchuyentb

#3 h Sep 19, 2010, 2:32 PM • 2 Y

Y by Adventure10, Mango247

truongtansang89 wrote:

$RHS \ge 3 \ge LHS$
D:

You are mistaken

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#4 h Sep 20, 2010, 6:07 AM • 2 Y

Y by Adventure10, Mango247

nguyenhtchuyentb wrote:

Let a,b,c be real posives such that a+b+c=3. Prove that
$\sqrt[3]{a} +\sqrt[3]{b} +\sqrt[3]{c} \geq ab+bc+ca$

Hello. It ís a easy problem. I have a nice problem.
Let a,b,c be real posives such that $a+b+c=3$ . Prove that
Click to reveal hidden text

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Ovchinnikov Denis

470 posts

Ovchinnikov Denis

#5 h Sep 20, 2010, 10:38 AM • 1 Y

Y by Adventure10

Lionel Messi wrote:

Hello. It ís a easy problem. I have a nice problem.

Can you post solution to "easy problem"?

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#6 h Sep 20, 2010, 11:07 AM • 2 Y

Y by Adventure10, Mango247

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kuing

#7 h Sep 22, 2010, 1:43 PM • 2 Y

Y by Adventure10, Mango247

nguyenhtchuyentb wrote:

Let a,b,c be real posives such that a+b+c=3. Prove that
$\sqrt[3]{a} +\sqrt[3]{b} +\sqrt[3]{c} \geq ab+bc+ca$

same condition, find all $k$ such that $a^k+b^k+c^k \ge ab+bc+ca$

maybe so hard ...

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#8 h Sep 22, 2010, 3:03 PM • 2 Y

Y by Adventure10, Mango247

I have found a proof for the problem as given, but it is far from elegant.

Use Lagrange multipliers to extremize $f(a,b,c) \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} - ab - bc - ca$ subject to $g(a,b,c) = a + b + c = 3.$

From this, we get $\frac{1}{3}a^{-\frac{2}{3}} - (b + c) + \lambda = \frac{1}{3}a^{-\frac{2}{3}} + a - 3 + \lambda = 0$ and analogous equations for $b$ and $c$ .

Now, differentiation will verify that $h(x) = \frac{1}{3}x^{-\frac{2}{3}} + x + 3$ has one minimum in $(0, 3)$ . This implies that at most two distinct values of $x$ will return the same value of $h(x)$ , and so at least two of $a, b, c$ are equal. Assume WLOG $a = b.$ We then have the equation $h(a) = h(3-2a),$ which has three solutions: $a \approx 0.03604612418654541, a \approx 1.453322263008721$ and $a = 1.$ Substitution will show that the minimum of $f(a,b,c) = 0$ when $a=b=c=1,$ and thus the original inequality is proved.

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Ovchinnikov Denis

470 posts

Ovchinnikov Denis

#9 h Sep 23, 2010, 4:52 AM • 2 Y

Y by Adventure10, Mango247

I don't understand

How $h(x)$ and $f(a,b,c)$ are related?
And why at least two of $a, b, c$ are same?

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#10 h Sep 23, 2010, 6:34 AM • 2 Y

Y by Adventure10, Mango247

I guess I should've written it out a little more clearly.

The method of Lagrange multipliers puts the extrema at the gradient of a function. That is, if we want to extremize $f(a,b,c) = LHS - RHS$ subject to $g(a,b,c)=a+b+c-3$ , we define $\Lambda(a,b,c,\lambda)=f(a,b,c) + \lambda \cdot (g(a,b,c) - 3)$ and we want to find all solutions that will make $\frac{\partial{\Lambda}}{\partial{a}}=\frac{\partial{\Lambda}}{\partial{b}}=\frac{\partial{\Lambda}}{\partial{c}}=\frac{\partial{\Lambda}}{\partial{\lambda}}=0.$
The last of these partial derivatives is the original constraint.
Now, this implies that, taking partial derivatives,
$\frac{1}{3}a^{-\frac{2}{3}}-(b+c)+\lambda =\frac{1}{3}a^{-\frac{2}{3}}+a-3+\lambda = 0 \\ \frac{1}{3}b^{-\frac{2}{3}}-(c+a)+\lambda =\frac{1}{3}b^{-\frac{2}{3}}+b-3+\lambda = 0 \\ \frac{1}{3}c^{-\frac{2}{3}}-(a+b)+\lambda =\frac{1}{3}c^{-\frac{2}{3}}+c-3+\lambda = 0.$
With the definition of $h(x) =\frac{1}{3}x^{-\frac{2}{3}}+x-3+\lambda$ (I apologize, that was a typo earlier; it should've been "minus," not "plus," though it doesn't change the differentiation at all. Anyway, I think this definition of $h$ is better now.), we rewrite that as $h(a) = h(b) = h(c) = 0.$
But, we note that $\frac{dh}{dx}=1-\frac{2}{9}x^{-\frac{5}{3}}$
and this has exactly one root in $(0, 3)$ ; this implies that over the interval, the function only changes direction once, and for any constant $k, h(x)=k$ has at most two distinct solutions in the interval.
Since $h(a)=h(b)=h(c),$ it must follow that at least two of $a,b,c$ have the same value.
From there there are only three cases to check, and each one of them will give $f(a,b,c) \ge 0,$ so the inequality is proved.

Of course, I think there is a much nicer solution to this one that makes use of a lot less calculus, but I haven't figured it out.

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Ovchinnikov Denis

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Ovchinnikov Denis

#11 h Sep 23, 2010, 7:11 AM • 2 Y

Y by Adventure10, Mango247

Thanks

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nguyenhtchuyentb

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nguyenhtchuyentb

#12 h Sep 24, 2010, 8:23 AM • 2 Y

Y by Adventure10, Mango247

Thank every body,

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nguyenhtchuyentb

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nguyenhtchuyentb

#13 h Sep 24, 2010, 8:34 AM • 2 Y

Y by Adventure10, Mango247

Let a,b,c,d be posive reals
Prove that:
$3(a+b+c+d) \le (ab+ac+ad+bc+bd+cd)(\sqrt{\frac{2}{3}(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad} +\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd})}$

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arqady

#14 h Sep 24, 2010, 9:38 AM • 1 Y

Y by Adventure10

nguyenhtchuyentb wrote:

Let a,b,c,d be posive reals
Prove that:
$3(a+b+c+d) \le (ab+ac+ad+bc+bd+cd)(\sqrt{\frac{2}{3}(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{ad} +\frac{1}{bc}+\frac{1}{bd}+\frac{1}{cd})}$

It's Maclaurin's inequality after following substitution: $a\rightarrow\frac{1}{a}$ , $b\rightarrow\frac{1}{b}$ , $c\rightarrow\frac{1}{c}$ and $d\rightarrow\frac{1}{d}$ .

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kuing

#15 h Oct 28, 2010, 8:22 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

nguyenhtchuyentb wrote:

Let a,b,c be real posives such that a+b+c=3. Prove that
$\sqrt[3]{a} +\sqrt[3]{b} +\sqrt[3]{c} \geq ab+bc+ca$

haha, at long last, I get a prove for it, excited!

here it is my prove:

$a,b,c \ge 0$ such that $a+b+c=3$ , prove that
$\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge ab + bc + ca.$
prove:
WOLG, assume $c = \min \left\{ {a,b,c} \right\}$ .
${\rm{Case}}\;1 : c < \frac{1}{6}$
easy to prove $\sqrt[3]{c} \ge 3c$ , then we get
$\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge \sqrt[3]{a} + \sqrt[3]{b} + 3c = \sqrt[3]{a} + \sqrt[3]{b} + c^2 + bc + ca,$ so we just need to prove that
$\sqrt[3]{a} + \sqrt[3]{b} \ge ab.$ setting $t = \sqrt[6]{{ab}}$ , by Am-Gm Inequality, we have
$\sqrt[3]{a} + \sqrt[3]{b} - ab \ge 2t - t^6 = t\left( {2 - t^5 } \right)\;{\rm{and}}\;0 \le t^5 \le \left( {\frac{{a + b}}{2}} \right)^{\frac{5}{3}} \le \left( {\frac{3}{2}} \right)^{\frac{5}{3}} ,$ but $8 - \left( {\frac{3}{2}} \right)^5 = \frac{{13}}{{32}} > 0$ gives $t\left( {2 - t^5 } \right) \ge 0$ , i.e. $\sqrt[3]{a} + \sqrt[3]{b} \ge ab$ .
so in this case, inequality is true;
${\rm{Case}}\;2:c \ge \frac{1}{6}$
we have
$\begin{align*} & \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge ab + bc + ca \\ \iff &\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} \ge \frac{{\left( {a + b + c} \right)^2 - \left( {a^2 + b^2 + c^2 } \right)}}{2} \\ \iff &2\left( {\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c}} \right) + a^2 + b^2 + c^2 \ge 9 \\ \iff &\sum\limits_{cyc} {\left( {2\sqrt[3]{a} + a^2 - \frac{{8a + 1}}{3}} \right)} \ge 0 \\ \iff &\sum\limits_{cyc} {\left( {\frac{1}{3}\left( {\sqrt[3]{a} - 1} \right)^2 \left( {3\sqrt[3]{{a^4 }} + 9\sqrt[3]{{a^2 }} + 4\sqrt[3]{a} + 6a - 1} \right)} \right)} \ge 0, \end{align*}$ obvious true because $a,b,c\ge\frac{1}{6}$ , so in this case, the inequality also true.
the prove is completed. done!

hope this prove is right, if you find some wrong, please tell me ...

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#16 h Oct 31, 2010, 11:52 AM • 2 Y

Y by Adventure10, Mango247

Wonderful!
I've verified Case 1, and though I haven't checked the last factorization in Case 2, it looks right.

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DC93

#17 h Dec 5, 2010, 11:32 AM • 2 Y

Y by Adventure10, Mango247

this problem isn't hard
notice that
$2(ab + bc + ca) = 9 - (a^2 + b^2 + c^2 )$
then we just need to prove
$\sum {a^2 + 2\sqrt[3]{a}} \ge 9$
It's not hard to prove

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kuing

#18 h Dec 5, 2010, 11:47 AM • 2 Y

Y by Adventure10, Mango247

DC93 wrote:

this problem isn't hard
notice that
$2(ab + bc + ca) = 9 - (a^2 + b^2 + c^2 )$
then we just need to prove
$\sum {a^2 + 2\sqrt[3]{a}} \ge 9$
It's not hard to prove

in my #15 Case2 can see this transformation, after this I use tangent line method, but not hold for all so I consider two cases to prove, have you got a more simpler solution for this?

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arqady

#19 h Dec 5, 2010, 5:44 PM • 1 Y

Y by Adventure10

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=336093

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kuing

#20 h Dec 6, 2010, 3:14 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=336093

but I don't think it's true, I have reply in the link.

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arqady

#21 h Dec 6, 2010, 5:33 AM • 2 Y

Y by Adventure10, Mango247

kuing wrote:

arqady wrote:

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=336093

but I don't think it's true, I have reply in the link.

I agree with you, kuing. My link is not relevant.

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