Really classical inequatily from canada

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shobber

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shobber

#1 h Mar 5, 2006, 2:50 AM • 3 Y

Y by mathematicsy, Adventure10, HWenslawski

Prove that for all positive real numbers $a$ , $b$ , and $c$ ,
$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c$
and determine when equality occurs.

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shobber

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shobber

#2 h Mar 5, 2006, 2:59 AM • 4 Y

Y by Adventure10, HWenslawski, Mango247, alexanderchew

By rearrangement: $a^3 \geq b^3 \geq c^3$ and $\frac{1}{bc} \geq \frac{1}{ac} \geq \frac{1}{ab}$ .

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Gibbenergy

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Gibbenergy

#3 h Mar 5, 2006, 9:12 PM • 7 Y

Y by xlzq, Frestho, Popslala, Adventure10, myh2910, Nabilos, kiyoras_2001

shobber wrote:

Prove that for all positive real numbers $a$ , $b$ , and $c$ ,
$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c$
and determine when equality occurs.

It isn't hard.

By AM-GM, we have
$\frac{a^3}{bc} +b+c \geq 3a$
Sum up and done

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pvthuan

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pvthuan

#4 h Mar 6, 2006, 3:09 AM • 3 Y

Y by Adventure10, rayfish, Nabilos

Yes, the inequality is simple applications of $x^2+y^2+z^2\geq xy+yz+zx$ for $a^2, b^2, c^2$ and $ab, bc,ca,$ .
We have
$a^4+b^4+c^4\geq a^2b^2+b^2c^2+c^2a^2\geq abc(a+b+c).$

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Davron

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Davron

#5 h Mar 7, 2006, 5:50 AM • 2 Y

Y by Adventure10, Mango247

Also you can prove the last one by the Muirheads Inequality...

Davron

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mssmath

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mssmath

#6 h Jan 15, 2014, 11:42 PM • 1 Y

Y by Adventure10

Now sum cyclically using AM-GM that
$\frac{a^3}{2bc}+\frac{b^3}{4ac}+\frac{c^3}{4ab}\ge a$

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chengchu

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chengchu

#7 h Jan 16, 2014, 4:31 AM • 3 Y

Y by Nabilos, Adventure10, Mango247

By $Cauchy-Schwarz$ and $AM-GM$ , we have:
$\sum\frac{a^3}{bc}\ge \frac{(a^2+b^2+c^2)^2}{3abc}\ge \frac{(ab+bc+ca)^2}{3abc}=a+b+c$

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sqing

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sqing

#8 h Jan 16, 2014, 6:10 AM • 2 Y

Y by Adventure10, Mango247

Very simple.
By AM-GM
$\sum( \frac{a^3}{bc} +b+c)\ge 3\sum a\iff \frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c$

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sayantanchakraborty

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sayantanchakraborty

#9 h Feb 28, 2014, 4:51 PM • 2 Y

Y by Adventure10, Mango247

Too easy.Using the fact that $\sum {x^2}\ge \sum{xy}$ we see that
$\sum{a^4} \ge \sum{a^2b^2} \ge \sum{a^2bc}$ .
Result follows by dividing both sides by $abc$ .

Bye...

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spikerboy

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spikerboy

#10 h Feb 28, 2014, 4:59 PM • 2 Y

Y by Adventure10, Mango247

It's trivial by chebychev, $\sum \frac{a^3}{bc} \ge\frac{a+b+c}{3} \sum \frac{a^2}{bc} \ge a+b+c$
I am learning inequalities and found chebychev very useful tool.

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math_science

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math_science

#11 h Jun 2, 2014, 9:03 PM • 2 Y

Y by Adventure10, Mango247

Inequality is equivalent to proving that $f(a,b,c) \geq 0$ where
$f(a,b,c)=a^4+b^4+c^4-a^2 b c-a b^2 c-a b c^2$ .
It is symmetric so we can take $a=c+x+y$ and $b=c+x$ for some $x,y \geq 0$ and then
$f(a,b,c)=f(c+x+y,c+x,c)=5 c^2 x^2+5 c^2 x y+5 c^2 y^2+6 c x^3+9 c x^2 y+11 c x y^2+4 c y^3+2 x^4+4 x^3 y+6 x^2 y^2+4 x y^3+y^4 \geq 0$
Done.

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arqady

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arqady

#12 h Apr 11, 2015, 7:27 PM • 1 Y

Y by Adventure10

shobber wrote:

Prove that for all positive real numbers $a$ , $b$ , and $c$ ,
$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c$
and determine when equality occurs.

The following inequality is also true.
Let $a$ , $b$ and $c$ be positive numbers such that $a^{15}+b^{15}+c^{15}=3$ . Prove that:
$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq3$

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VadimM

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VadimM

#13 h Apr 11, 2015, 7:46 PM • 3 Y

Y by MATH1945, Adventure10, Mango247

shobber wrote:

Prove that for all positive real numbers $a$ , $b$ , and $c$ ,
$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c$
and determine when equality occurs.

$\iff a^4+b^4+c^4\geq a^2b^2+c^2a^2+b^2c^2\geq abc(a+b+c)$ very simple

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aditya21

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aditya21

#14 h Apr 12, 2015, 6:48 PM • 2 Y

Y by Adventure10, Mango247

shobber wrote:

Prove that for all positive real numbers $a$ , $b$ , and $c$ ,
$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a+b+c$
and determine when equality occurs.

alternatively by tchebytcheff we have
$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab}\ge [\frac{(a+b+c)}{3}](\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab})$
but by cauchy
$[\frac{(a+b+c)}{3}](\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab})\ge a+b+c$

as $(a+b+c)^2\ge 3(ab+bc+ca)$

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AdithyaBhaskar

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AdithyaBhaskar

#15 h Jul 29, 2015, 4:49 PM • 2 Y

Y by Adventure10, Mango247

Here is another solution:
Note that the inequality is homogeneous in $a,b,c$ so we let $abc=1$ . The first observation is that, by AM-GM, $a+b+c \geq 3\sqrt[3]{abc} = 3$ . Further, the given inequality reduces to $a^4 + b^4 + c^4 \geq a + b + c$ . This is easy, as by the AM-QM (or the power mean) inequality, we have $a^4 + b^4 + c^4 \geq 4 \cdot (\frac{a+b+c}{4})^4 = (a+b+c) (\frac{a+b+c}{3})^3 \geq a+b+c$ , and we are done.

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ryanbear

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ryanbear

#71 h Oct 9, 2023, 3:34 PM

Y by

Multiply on $abc$ on both sides to get $a^4+b^4+c^4 \ge a^2bc+a^2bc+abc^2$
Because $(4,0,0) > (2,1,1)$ , the inequality is true

This post has been edited 1 time. Last edited by ryanbear, Oct 9, 2023, 3:35 PM

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CrazyInMath

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CrazyInMath

#72 h Oct 9, 2023, 4:41 PM

Y by

One-Liner
$\sum \frac{a^3}{bc}=\sum\frac{\frac{a^3}{bc}+\frac{c^3}{ab}}{2}\geq\sum\frac{ac}{b}=\sum\frac{\frac{ac}{b}+\frac{ab}{c}}{2}\geq\sum a$

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adityaguharoy

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adityaguharoy

#73 h Oct 9, 2023, 10:26 PM

Y by

CrazyInMath wrote:

One-Liner
$\sum \frac{a^3}{bc}=\sum\frac{\frac{a^3}{bc}+\frac{c^3}{ab}}{2}\geq\sum\frac{ac}{b}=\sum\frac{\frac{ac}{b}+\frac{ab}{c}}{2}\geq\sum a$

This is nice, but most solutions above are one-liner.
Here is a one0line rephrasement of the most commonly discussed solution.

By rearrangement $\frac{a^3}{bc} +\frac{b^3}{ca}+\frac{c^3}{ab} = \frac{a(a^2)}{bc}+\frac{b(b^2)}{ca}+\frac{c(c^2)}{ab} \ge \frac{a(bc)}{bc} + \frac{b(ca)}{ca}+\frac{c(ab)}{ab} = a+b+c.$

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gracemoon124

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gracemoon124

#74 h Dec 8, 2023, 5:06 AM

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Multiply both sides by $abc$ , so it suffices to prove $a^4+b^4+c^4\ge a^2bc+ab^2c+abc^2$ .

Notice how $a^4+a^4+b^4+c^4\ge 4a^2bc$ by AM-GM. Summing this with its analogous inequalities proves the result, and equality occurs when $a=b=c$ .

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ytChen

1092 posts

ytChen

#75 h Jan 7, 2024, 6:51 PM

Y by

sqing wrote:

Prove that for all positive real numbers $a$ , $b$ , and $c$ ,
$\frac{a^3}{bc} + \frac{b^2c^2}{a^2} \geq ab+c$

Remark. The claimed inequality $\frac{a^3}{bc} + \frac{b^2c^2}{a^2} \geq ab+c$ is not true for all positive real numbers $a,b,c$ . For instance, if $\left(a,b,c\right)=\left(1,\frac{1}{2},3\right)$ , then
$\begin{align*}\frac{a^3}{bc} + \frac{b^2c^2}{a^2}-( ab+c) =&\frac{2}{3}+\frac{9}{4}-\frac{1}{2}-3\\ =&-\frac{7}{12}<0. \end{align*}$

This post has been edited 1 time. Last edited by ytChen, Jan 7, 2024, 6:53 PM
Reason: Typo

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AshAuktober

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AshAuktober

#76 h Feb 8, 2024, 3:52 PM • 1 Y

Y by lelouchvigeo

Observe that $(4, 0, 0) \succ (2,1,1)$ , so from Muirhead, $a^4 + b^4 + c^4 \ge a^2bc + b^2ca + c^2ab$ , and dividing by $abc$ gives the desired result. Since Muirhead can be achieved from various applications of AM-GM, equality holds iff $a = b = c$ .

This post has been edited 1 time. Last edited by AshAuktober, Feb 8, 2024, 4:08 PM

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Shreyasharma

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Shreyasharma

#77 h Feb 8, 2024, 6:25 PM

Y by

Note that upon expansion we have,
$\begin{align*} \sum a^5bc \geq \sum a^3b^2c^2\\ \end{align*}$ Then homogenize so that $abc = 1$ . Then we wish to show,
$\begin{align*} \sum a^4 &\geq \sum a\\ \iff \sum a^4-a &\geq 0 \end{align*}$ Now define $f(x) = x^4 - x$ . We may check,
$\begin{align*} f''(x) = 12x^2 \end{align*}$ and hence $f$ is convex everywhere. However then by Jensen's we may check,
$\begin{align*} f(a) + f(b) + f(c) \geq 3\cdot f\left(\frac{a+b+c}{3} \right) \end{align*}$ Note that by AM-GM
$\begin{align*} \frac{a+b+c}{3} \geq 1 \end{align*}$ Finally we may check,
$\begin{align*} f([x \geq 1]) \geq f(1) = 0 \end{align*}$ Indeed note that this holds as,
$\begin{align*} x^4 - x \geq 0 \iff x(x^3 - 1) \geq 0 \iff x \in (-\infty, 0) \cup (1, \infty) \end{align*}$ Thus we may finish as,
$\begin{align*} \sum f(a) \geq 3 \cdot f \left( \frac{a+b+c}{3} \right) \geq 3 \cdot f(1) \geq 0 \end{align*}$

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shanelin-sigma

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shanelin-sigma

#78 h Jun 18, 2024, 5:43 AM

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From Cauchy Inequality, $(\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab})(b+c+a)(c+a+b) \geq (a+b+c)^3$ Divide by $(a+b+c)^2$ at both sides and QED
(Equality occurs when $a=b=c$ )

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anduran

466 posts

anduran

#79 h Jun 18, 2024, 6:27 AM

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shanelin-sigma wrote:

From Cauchy Inequality, $(\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab})(b+c+a)(c+a+b) \geq (a+b+c)^3$ Divide by $(a+b+c)^2$ at both sides and QED
(Equality occurs when $a=b=c$ )

This is Holder's inequality.

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anduran

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anduran

#80 h Jun 18, 2024, 6:30 AM

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Starting from the already stated
$a^4 + b^4 + c^4 \geq abc(a+b+c),$ We can note that
$a^4 + b^4 + c^4 \geq a^2b^2 + b^2c^2 + c^2a^2 \geq (ab)(bc) + (bc)(ca) + (ca)(ab) = abc(a+b+c)$

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anduran

466 posts

anduran

#81 h Jun 19, 2024, 1:25 AM

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We can also use the Holder
$\left(\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\right)(bc + ca + ab)(1 + 1 + 1) \geq (a + b + c) ^3$ So it remains to prove
$(a+b+c)^3 \geq 3(a+b+c)(ab+bc+ca)$ Which is obvious since $(a+b+c)^2 \geq 3(ab+bc+ca)$

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saumonx07

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saumonx07

#82 h Oct 17, 2024, 10:43 AM

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Rewrite as $a^4 + b^4 + c^4$ $\geq$ $a^2bc + b^2ac + c^2ab$
By AM-GM,
$a^4 + a^4 + b^4 + c^4$ $\geq$ $4a^2bc$
Equality when $a$ = $b$ = $c$

This post has been edited 1 time. Last edited by saumonx07, Oct 17, 2024, 10:45 AM
Reason: equality

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cappucher

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cappucher

#83 h Dec 22, 2024, 12:50 AM

Y by

By AM-GM, we have

$\frac{a^3}{bc} + b + c \geq 3a$
Applying a cyclic sum, we have

$\sum_{\text{cyc}} \left( \frac{a^3}{bc} + b + c \right) \geq 3(a + b + c)$ $\sum_{\text{cyc}} \frac{a^3}{bc} + 2(a + b + c) \geq 3(a + b + c)$ $\sum_{\text{cyc}} \frac{a^3}{bc} \geq a + b + c$
as desired.

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Jupiterballs

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Jupiterballs

#84 h Dec 26, 2024, 6:39 AM

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trivial by muirheads,
As $(4,0,0) \succ (2,1,1)$
$a^4 + b^4 + c^4 \ge ab^2c + abc^2 + a^2bc$
which gets,
$a^4 + b^4 + c^4 \ge abc(a + b + c)$
$\frac{a^3}{bc} + \frac{b^3}{ac} + \frac{c^3}{ab} \ge (a + b + c)$
Hence proved

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Tony_stark0094

44 posts

Tony_stark0094

#85 h Mar 31, 2025, 4:19 PM

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AM-GM to get
$\sum (\frac{a^3}{bc} +b+c)\geq \sum 3a$ and the required inequality follows

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