min of (a^3 + b^3 + c^3)/abc where a,b,c sidelengths of right triangle

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min of (a^3 + b^3 + c^3)/abc where a,b,c sidelengths of right triangle

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Source: Thailand Mathematical Olympiad 2017 day 2 p8

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#1 h Jan 30, 2020, 9:51 AM • 1 Y

Y by Adventure10

Let $a, b, c$ be side lengths of a right triangle. Determine the minimum possible value of $\frac{a^3 + b^3 + c^3}{abc}$ .

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#3 h Jan 30, 2020, 10:04 AM • 2 Y

Y by Adventure10, Mango247

parmenides51 wrote:

Let $a, b, c$ be side lengths of a right triangle. Determine the minimum possible value of $\frac{a^3 + b^3 + c^3}{abc}$ .

$minimum?$
$\frac{a^3 + b^3 + c^3}{abc}\geq \frac{3abc}{abc}=3$

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sqing

#4 h Jan 30, 2020, 12:06 PM • 3 Y

Y by Adventure10, Mango247, Mango247

NO.
$\frac{a^3 + b^3 + c^3}{abc}\geq 2+\sqrt{2}.$

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#5 h Jan 30, 2020, 12:24 PM • 2 Y

Y by Adventure10, ehuseyinyigit

parmenides51 wrote:

Let $a, b, c$ be side lengths of a right triangle. Determine the minimum possible value of $\frac{a^3 + b^3 + c^3}{abc}$ .

Suppose that $c$ is the hypothenuse, so $c^2=a^2+b^2$ . From Holder $2(a^3+b^3)^2 \geq (a^2+b^2)^3$ , so $a^3+b^3 \geq \frac{c^3}{\sqrt{2}}$ . Also $ab \leq \frac{1}{2}(a^2+b^2)=\frac{1}{2}c^2$ Therefore:

$\frac{a^3+b^3+c^3}{abc} \geq \frac{\frac{c^3}{\sqrt{2}}+c^3}{\frac{c^3}{2}}= 2+\sqrt{2}$

Equality when $a =b=\frac{c}{\sqrt{2}}$ .

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#6 h Jan 30, 2020, 12:28 PM • 1 Y

Y by Adventure10

Assume wlog $a^2=b^2+c^2$
Then
AM-GM: $a^2=b^2+c^2\ge 2\sqrt{bc}$ , power mean inequality $b^3+c^3\ge \frac{1}{\sqrt2}\cdot a^3$
$(b-c)^2+\frac{1}{\sqrt2}\cdot a^2\ge \sqrt2bc$
$a(b-c)^2+\frac{1}{\sqrt2}\cdot a^3\ge \sqrt2abc$
$a(b^2-2bc+c^2)+b^3+c^3\ge \sqrt2abc$
$\frac{a^3 + b^3 + c^3}{abc}\geq 2+\sqrt{2}.$
Equality iff $b=c,a=\sqrt2 b$

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sqing

#7 h Jan 30, 2020, 12:43 PM • 2 Y

Y by Adventure10, Tomas_Klevas

Let $a,b,c$ be three positive real numbers such that $a^2=b^2+c^2$ . Prove that $a^3+b^3+c^3\ge(2+\sqrt{2})abc.$
$a^3+b^3+c^3=\left(1-\frac{\sqrt{2}}{4}\right)a^3+\frac{\sqrt{2}}{4}a^3+b^3+c^3\geq2\left(1-\frac{\sqrt{2}}{4}\right)abc+\frac{3\sqrt{2}}{2}abc=(2+\sqrt{2})abc.$ Equality holds when $a=\sqrt{2}b=\sqrt{2}c.$

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sqing

#8 h Jan 30, 2020, 1:07 PM • 1 Y

Y by Adventure10

Let $a, b, c$ be side lengths of a right triangle. Prove that $\frac{a^4 + b^4 + c^4}{abc(a+b+c)}\geq3(\sqrt 2-1).$

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#9 h Jan 30, 2020, 1:43 PM • 2 Y

Y by Adventure10, Mango247

WLOG let $a> b\geq c$
$a+b\leq \sqrt{2(a^2+b^2)}=\sqrt2 c$
So, $a+b+c\leq (\sqrt{2}+1)c$
$ab\leq \frac{a^2+b^2}{2}=\frac{c^2}{2}$
$a^4+b^4+c^4\geq \frac{(a^2+b^2)^2}{2}+c^4=\frac{3}{2}c^4$
Dividing the third by the first 2, we get the required bound

This post has been edited 2 times. Last edited by Math-wiz, Jan 30, 2020, 2:08 PM

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#10 h Jan 30, 2020, 1:56 PM • 1 Y

Y by Adventure10

Thanks.
Let $a, b, c$ be side lengths of a right triangle. Prove that $\frac{a^5 + b^5 + c^5}{abc(ab+bc+ca)}\geq\frac{8(\sqrt 2-1)}{7}.$

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#11 h Jan 30, 2020, 2:26 PM • 1 Y

Y by Adventure10

Let $c$ be the hypotenuse
$a+b\leq \sqrt2 c$ , as proved earlier
$ab\leq\frac{c^2}{2}$ as proved earlier
By power mean, $\sqrt[5]{\frac{a^5+b^5}{2}}\geq \sqrt{\frac{a^2+b^2}{2}}$ , so $a^5+b^5\geq \frac{c^5}{2\sqrt2}$ So, $LHS=\frac{(a^5+b^5)+c^5}{(ab)(c)(c(a+b)+ab)}$
Substituting the bounds proved above, we get the required bound

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sqing

#12 h Jan 31, 2020, 3:26 AM • 1 Y

Y by Adventure10

Let $a,b,c$ be positive numbers such that $a^2=b^2+c^2$ . Prove that $\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 5+3\sqrt{2}$ $a^3+b^3+c^3\ge(2+\sqrt{2})abc$ .
$a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)\ge(2+3\sqrt{2})abc$ .
$a^3+b^3+c^3\ge \frac{2\sqrt{2}+1}{7}[a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)]$ https://artofproblemsolving.com/community/c6h511724p2873311
https://artofproblemsolving.com/community/c6h608170p3615254
https://artofproblemsolving.com/community/c6h608170p3615254
https://artofproblemsolving.com/community/c6h2871334p25507389
Thanks.

This post has been edited 1 time. Last edited by sqing, Jul 16, 2022, 9:49 AM

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#13 h Apr 27, 2020, 8:31 AM

Y by

Math-wiz wrote:

WLOG let $a> b\geq c$
$a+b\leq \sqrt{2(a^2+b^2)}=\sqrt2 c$
So, $a+b+c\leq (\sqrt{2}+1)c$
$ab\leq \frac{a^2+b^2}{2}=\frac{c^2}{2}$
$a^4+b^4+c^4\geq \frac{(a^2+b^2)^2}{2}+c^4=\frac{3}{2}c^4$
Dividing the third by the first 2, we get the required bound

Very nice ,but i seems is WLOG let $c> a\geq b$

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#14 h May 30, 2024, 1:25 PM

Y by

Suppose that $c$ is the hypotenuse and $x$ is an angle between $c$ and $b$ . Thus,
$c \sin(x) = a \quad \text{and} \quad c \cos(x) = b.$ Then, substitute to find the minimum of
$\frac{(\sin(x))^3 + (\cos(x))^3 + 1}{\sin(x)\cos(x)}.$ Consider
$(\sin(x))^3 + (\cos(x))^3 + 1 = (\sin(x) + \cos(x))(1 - \sin(x)\cos(x)) + 1.$ This yields
$(\sin(x))^3 + (\cos(x))^3 + 1 \geq \sqrt{2} \left(\frac{1}{2}\right) + 1.$ And
$\sin(2x) + \frac{\sqrt{2}}{2} \sin(2x) \leq 1 + \frac{1}{\sqrt{2}}.$ Thus,
$(\sin(x))^3 + (\cos(x))^3 + 1 \geq \sin(2x) + \frac{\sqrt{2}}{2} \sin(2x).$ Rewrite $\sin(2x)$ as $2\sin(x)\cos(x)$ :
$\frac{(\sin(x))^3 + (\cos(x))^3 + 1}{\sin(x)\cos(x)} \geq 2 + \sqrt{2}.$ Equality holds when $a\sqrt{2} = b\sqrt{2} = c$ .

This post has been edited 6 times. Last edited by chokefern, Jun 14, 2024, 1:15 AM
Reason: Typo

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