Japan 1997 inequality

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hxtung

317 posts

hxtung

#1 h Jul 27, 2003, 9:18 AM • 5 Y

Y by smy2012, Adventure10, ehuseyinyigit, Mango247, and 1 other user

Prove that

$\frac{\left(b+c-a\right)^{2}}{\left(b+c\right)^{2}+a^{2}}+\frac{\left(c+a-b\right)^{2}}{\left(c+a\right)^{2}+b^{2}}+\frac{\left(a+b-c\right)^{2}}{\left(a+b\right)^{2}+c^{2}}\geq\frac35$

for any positive real numbers $a$ , $b$ , $c$ .

This post has been edited 1 time. Last edited by hxtung, Jul 29, 2003, 3:00 AM

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Lagrangia

1326 posts

Lagrangia

#2 h Jul 27, 2003, 12:04 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

okay here is a solution:

sum((2ab+2ac)/(a^2+b^2+c^2+2bc))<=24/5 (Note that the sum is sym).
lets denote s=a^2+b^2+c^2 for easy calculus!

5s^2(sum(ab))+10s(sum(a^2bc))+20(sum(a^3b^2c))<=6s^3+6s^2(sum(ab))+12s(sum(a^2bc))+48a^2b^2c^2 (all sums are sym)
which after some simplifacations and calculus we get:
6s^3+s^2sum(ab)+2s(sum(a^2bc)+8sum(a^2b^2c^2)>=10s(sum(a^2bc))+20(sum(a^3b^2c). (sums are sym)
we now multiply out the powers of s=a^2+b^2+c^2 and get:
sym sum 3a^6+2a^5b-2ab^2+3abc+2a^3b^3-12a^3b^2c+4a^2b^2c^2>=0
now using Schur's inequality after multiplication with 4abc:
sym sum 4abc-8a^3b^2c+4a^2b^2c^2>=0 and this reduces our claim to

sym sum 3a^6+2a^5b-2ab^2-abc+2a^3b^3-4a^3b^2c>=0.

now we have a sum of four expressions which are positive by weighted AM-GM:

2sym sum(2a^6+b^6)/3-ab^2>=0
sym sum(4a^6+b^6+c^6)/6-abc>=0
2 sym sum (2a^3b^3+c^3a^3)/3-a^3b^2c>=0
2 sys sum (2a^5b+a^5c+ab^5+ac^5)/6-a^3b^2c>=0

equality is when a=b=c ( we see this from Shur's ineq and AM-GM)

cheers hope this is good!

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iandrei

138 posts

iandrei

#3 h Jul 27, 2003, 2:43 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

I think I have an interesting solution for your problem . Here goes :
(all sums are cyclic , not symmetric)

sum (b+c-a)^2/[(b+c)^2 + a^2] >= 3/5 <=>
S = sum a(b+c)/[(b+c)^2 + a^2] <= 6/5 , after a few trivial computations

Now , (b+c)^2 + a^2 >= (a+2b+2c)^2/5 , by cauchy applied to the numbers (b+c)/2 , (b+c)/2 , (b+c)/2 , (b+c)/2 , a - we apply it in this manner in order to have equality .

This means that S <= sum 5a(b+c)/(a+2b+2c)^2 = S'

We prove that S' <= 6/5 . Assume that a+b+c=1 since the inequality is homogenous , and you end up with :

sum a(1-a)/(2-a)^2 <= 6/25

Consider the function f(x) = (x-x^2)/(2-x)^2 . If it is concave on (0,1) we are done . But this is true by a trivial computation (show that f''(x)<0 on (0,1)) . So we can apply Jensen's inequality and we are done .

Voila , no horrendous computations !

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Lagrangia

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Lagrangia

#4 h Jul 27, 2003, 4:15 PM • 4 Y

Y by Adventure10, Mango247, ehuseyinyigit, and 1 other user

well my solution is the so called "official solution" that was given at the contest! cheers!

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hxtung

317 posts

hxtung

#5 h Jul 28, 2003, 8:27 AM • 4 Y

Y by Adventure10, Mango247, EntropiaAwake, and 1 other user

Hey! every one
This my solution.I think, it is nice because ....
Put S=(b + c - a)2/((b+c)2+a2) +(a + c - b)2/((a+c)2+b2) + (b + a - c)2/((b+a)2+c2)
WLOG, We suppose a + b + c=1.(Because, If a + b+ c = x, we put a1=a/x, b1=b/x, c1=c/x
then inequality equivalent S >= 3/5 with a1+ b1 + c1=1 )
We have S=(1-2a)2/(a2 + (1-a)2) + (1-2b)2/(b2 + (1-b)2) + (1-2c)2/(c2 + (1-c)2)
We'll prove (1-2a)2/(a2 + (1-a)2)>= (23-54a)/25 (*).
Really multiplying 25((a2 + (1-a)2), we get 2(3a-1)2(6a+1) >=0, then (*) is true.
Applying (*), then we have (1-2b)2/(b2 + (1-b)2)>= (23-54b)/25 , (1-2c)2/(c2 + (1-c)2)>= (23-54c)/25
Then S>= (23.3 - 54(a + b +c))/25 = 3/5.
It's very good???

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alekk

2620 posts

alekk

#6 h Aug 8, 2003, 1:36 PM • 4 Y

Y by Adventure10, ehuseyinyigit, Mango247, and 1 other user

hey, how did you find 23/25 and -54/25??
(1-2a)2/(a2 + (1-a)2)>= (23-54a)/25

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hxtung

317 posts

hxtung

#7 h Aug 9, 2003, 3:01 AM • 4 Y

Y by smy2012, Adventure10, Mango247, and 1 other user

Hey, see my soln for poland 1996 inequality

[Moderator edit: Also compare with the USAMO 2003 inequality.]

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harazi

5526 posts

harazi

#8 h Jan 4, 2004, 11:35 AM • 4 Y

Y by Adventure10, Mango247, ehuseyinyigit, and 1 other user

I also have a short solution. Put x=(b+c)/a, y=(c+a)/b and z=(a+b)/c. Apply Cauchy to find that the LHS is sum (x-1)^2/(x^2+1)>=(x+y+z-3)^2/(sum x^2+3). This becomes (x+y+z)^2-15(x+y+z)+18+3(xy+yz+zx)>=0.But xy+yz+zx>=2(x+y+z) (since this one reduces to (a+b-c)(b+c-a)(c+a-b)<=abc), so it is enough to prove that (sum x)^2-9sumx+18>=0 and it is true since sumx>=6.

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Mildorf

785 posts

Mildorf

#9 h Jul 8, 2005, 3:28 AM • 5 Y

Y by smy2012, Adventure10, Mango247, ehuseyinyigit, and 1 other user

alekk wrote:

hey, how did you find 23/25 and -54/25??
(1-2a)2/(a2 + (1-a)2)>= (23-54a)/25

I think my solution basically employs the same idea. First, say $a+b+c=3$ so that equality will hold at $a=b=c=1$ . Essentially, one conjectures that

$\frac{(b+c-a)^2}{(b+c)^2+a^2} = \frac{(3-2a)^2}{(3-a)^2+a^2} \ge \frac{1}{5} + ka - k$

for some $k$ . As the equation is parametrically contrived to yield equality where $a=1$ , we need the $k$ for which there is a double root. That can be solved for by equating derivatives, etc. and you get $k = \frac{-18}{25}$ . Factoring for this $k$ , we see that the inequality is always true. Then add the three expressions to finish.

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Viet Math

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Viet Math

#10 h Jul 8, 2005, 6:13 AM • 2 Y

Y by Adventure10 and 1 other user

Nice but unnature

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robinho

36 posts

robinho

#11 h Jan 12, 2006, 2:16 AM • 5 Y

Y by Adventure10, Mango247, Mango247, Mango247, and 1 other user

If you want prove it, you should use SOS inequality

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tranthanhnam

420 posts

tranthanhnam

#12 h Jan 13, 2006, 3:18 AM • 3 Y

Y by Mathskidd, Adventure10, and 1 other user

$\frac{(b+c-a)^2}{(b+c)^2+a^2}+ \frac{(a+c-b)^2}{(a+c)^2+b^2}+\frac{(b+a-c)^2}{(b+a)^2+c^2} \geq 3/5$ (1)
let $a+b+c=3$ then (1) $\leq >$
$\frac{(3-2a)^2}{(3-a)^2+a^2} + \frac{(3-2b)^2}{(3-b)^2+b^2} + \frac{(3-2c)^2}{(3-c)^2+c^2} \geq 3/5$
now , it's easy

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HTA

339 posts

HTA

#13 h Apr 22, 2007, 2:38 AM • 4 Y

Y by Adventure10, Mango247, vrondoS, and 1 other user

we may assume that $a+b+c=2$ , the inequality becomes
$\sum\frac{(a-1)^{2}}{1+(a-1)^{2}}\geq \frac{3}{10}$
$\leftrightarrow \sum \frac{1}{1+(a-1)^{2}}\leq \frac{27}{10}$
now we denote $x=a-1;y=b-1;z=c-1$
the inequality becomes $\sum \frac{1}{1+x^{2}}\leq \frac{27}{10}$ with $x+y+z=1$ and $x,y,z <1$
by the fact $27(1-3t)^{2}(4-3t) \geq 0$ we obtain $\frac{1}{1+x^{2}}\leq \frac{27}{50}(2-x)$
now we easily get the desire result

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me@home

2349 posts

me@home

#14 h May 26, 2007, 11:44 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Mildorf wrote:

alekk wrote:

hey, how did you find 23/25 and -54/25??
(1-2a)2/(a2 + (1-a)2)>= (23-54a)/25

I think my solution basically employs the same idea. First, say $a+b+c=3$ so that equality will hold at $a=b=c=1$ . Essentially, one conjectures that

$\frac{(b+c-a)^{2}}{(b+c)^{2}+a^{2}}= \frac{(3-2a)^{2}}{(3-a)^{2}+a^{2}}\ge \frac{1}{5}+ka-k$

for some $k$ . As the equation is parametrically contrived to yield equality where $a=1$ , we need the $k$ for which there is a double root. That can be solved for by equating derivatives, etc. and you get $k = \frac{-18}{25}$ . Factoring for this $k$ , we see that the inequality is always true. Then add the three expressions to finish.

I tried this one with isolated fudging of the form $\frac{a^{p}}{a^{p}+b^{p}+c^{p}}$ and the exponent is the same as your k... coincidence?

But isolated fudging doesn't work so that's why I requested a link to here

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nealth

303 posts

nealth

#15 h Jul 10, 2007, 6:32 PM • 3 Y

Y by Adventure10, ehuseyinyigit, and 1 other user

There is actually a short solution:
$\frac{2a(b+c)}{(b+c)^{2}+a^{2}}=\frac{2a(b+c)}{\frac{3}{4}(b+c)^{2}+\frac{1}{4}(b+c)^{2}+a^{2}}\le \frac{2a(b+c)}{\frac{3}{4}(b+c)^{2}+(b+c)a}=\frac{2a}{\frac{3}{4}(b+c)+a}$
Now if we assume $a \le b \le c$ then $\frac{1}{\frac{3}{4}(b+c)+a}\ge \frac{1}{\frac{3}{4}(a+c)+b}\ge \frac{1}{\frac{3}{4}(b+a)+c}$
So by rearrangement, we deduce that this cyclic sum is $\le 12/5$
The rest follows easily.

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Nguyenhuyen_AG

3302 posts

Nguyenhuyen_AG

#63 h Mar 29, 2020, 3:53 AM

Y by

SBM wrote:

Better: https://artofproblemsolving.com/community/u493456h1939876p14491686

Your proof is #20

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Grizzy

920 posts

Grizzy

#64 h Dec 31, 2020, 4:19 AM

Y by

Since the inequality is homogeneous, we can assume the condition $a+b+c=1$ and scale up later. The inequality then becomes

$\begin{align*} \sum \frac{(1-2a)^2}{a^2 + (1-a)^2} \ge \frac{3}{5} \end{align*}$
which rearranges to

$\begin{align*} \frac{27}{5} \ge \sum \frac{1}{2a^2 - 2a + 1}. \end{align*}$
By the tangent line trick, we derive the inequality

$\frac{1}{2a^2-2a+1} \le \frac{9}{5} + \frac{54}{25}\left(a - \frac{1}{3}\right) \Longleftrightarrow 2(3a-1)^2(6a+1) \ge 0.$
Summing this cyclicly yields the desired inequality. $\square$

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Tafi_ak

309 posts

Tafi_ak

#65 h Jan 5, 2022, 10:19 PM

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Since the inequality is homogeneous therefore WLOG we assume $a+b+c=3$ . Now the inequality becomes
$\begin{align*} \sum_{cyc}\frac{(3-2a)^2}{a^2+(3-a)^2}\geq \frac{3}{5} \end{align*}$ Consider the function $f(x)=\frac{(3-2x)^2}{x^2+(3-x)^2}$ . So we have to prove that $f(a)+f(b)+f(c)\geq \frac{3}{5}$ Notice that (Tangent line trick) $f(x)\geq \frac{23-18x}{25}\Longleftrightarrow \frac{(x-1)^2(2x+1)}{2x^2-6x+9}\geq 0$ is true (the denominator $2x^2-6x+9>0$ because the coefficient of $x^2$ is positive and least value is $\left(\frac{3}{2}, \frac{9}{2}\right)$ .) for all value of positive $x$ . Now just summing up them we get our desired result.

This post has been edited 1 time. Last edited by Tafi_ak, Jan 5, 2022, 10:21 PM
Reason: big brackets

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bryanguo

1032 posts

bryanguo

#66 h Jun 30, 2022, 12:24 AM • 2 Y

Y by channing421, Mango247

Same as everyone else. Nice application of TLT.

Since both sides of the inequality are of same degree, the inequality is homogeneous. Since the domain is positive reals as well, along with the previous fact, we can assume $a+b+c=3.$ Our inequality we want to prove rewrites to $\sum_{\text{cyc}} \frac{(3-2a)^2}{a^2+(3-a)^2} \geq \frac{3}{5}.$ Recall the Tangent Line Trick, which is if we fix $m=\tfrac{a_1+\cdots+a_n}{n},$ and if $f$ is nonconvex, then if $f(x) \geq f(m)+f'(m)(x-m)$ holds for all $x,$ summing cyclically produces the desired result. In our case, consider the function $f(x)=\tfrac{(3-2x)^2}{x^2+(3-x)^2},$ then we want to prove $f(a)+f(b)+f(c) \geq \tfrac{3}{5}.$ Since the second derivative of $f$ is negative, the function is nonconvex. By applying the Tangent Line Trick, $\frac{(3-2a)^2}{(3-a)^2+a^2} \geq \frac{1}{5}-\frac{18}{25}(a-1) \implies$ $\frac{18}{25}(a-1)-\frac{1}{5}+\frac{(3-2a)^2}{(3-a)^2+a^2} = \frac{18(a-1)^2(2a+1)}{25(2a^2-6a+9)} \geq 0,$ of which the last inequality is clearly true because $a$ is a positive real. Thus by Tangent Line Trick, summing cyclically implies the result.

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sqing

41424 posts

sqing

#67 h Jun 30, 2022, 12:44 AM

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Prove that
$\frac{(b+c-5a)^2}{(b+c)^2+5a^2}+\frac{(c+a-5b)^2}{(c+a)^2+5b^2}+\frac{(a+b-5c)^2}{(a+b)^2+5c^2}\geq 3$ for any positive real numbers $a$ , $b$ , $c$ .

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arqady

30171 posts

arqady

#68 h Jun 30, 2022, 4:18 AM

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sqing wrote:

Prove that
$\frac{(b+c-5a)^2}{(b+c)^2+5a^2}+\frac{(c+a-5b)^2}{(c+a)^2+5b^2}+\frac{(a+b-5c)^2}{(a+b)^2+5c^2}\geq 3$ for any positive real numbers $a$ , $b$ , $c$ .

It's true for any reals $a$ , $b$ and $c$ such that $\prod\limits_{cyc}((a+b)^2+5c^2)\neq0.$

This post has been edited 1 time. Last edited by arqady, Jun 30, 2022, 4:33 AM

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sqing

41424 posts

sqing

#69 h Jun 30, 2022, 6:21 AM • 2 Y

Y by Mango247, Mango247

Prove that
$\frac{a\left(a+b-c\right)^{2}}{\left(a+b\right)^{3}+c^{3}}+\frac{b\left(b+c-a\right)^{2}}{\left(b+c\right)^{3}+a^{3}}+\frac{c\left(c+a-b\right)^{2}}{\left(c+a\right)^{3}+b^{3}}\geq\frac1{3}$ $\frac{\left(b+c-a\right)^{3}}{\left(b+c\right)^{3}+a^{3}}+\frac{\left(c+a-b\right)^{3}}{\left(c+a\right)^{3}+b^{3}}+\frac{\left(a+b-c\right)^{3}}{\left(a+b\right)^{3}+c^{3}}\geq\frac1{3}$ $\frac{a\left(a+b-c\right)^{2}}{\left(a+b\right)^{3}+2c^{3}}+\frac{b\left(b+c-a\right)^{2}}{\left(b+c\right)^{3}+2a^{3}}+\frac{c\left(c+a-b\right)^{2}}{\left(c+a\right)^{3}+2b^{3}}\geq\frac3{10}$ $\frac{\left(b+c-a\right)^{3}}{\left(b+c\right)^{3}+2a^{3}}+\frac{\left(c+a-b\right)^{3}}{\left(c+a\right)^{3}+2b^{3}}+\frac{\left(a+b-c\right)^{3}}{\left(a+b\right)^{3}+2c^{3}}\geq\frac3{10}$ for any positive real numbers $a$ , $b$ , $c$ .
Thanks.

This post has been edited 2 times. Last edited by sqing, Jul 13, 2022, 2:37 AM

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HamstPan38825

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HamstPan38825

#70 h Mar 19, 2023, 5:37 PM

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De-homogenize with $a+b+c = 1$ . Now, notice that $\frac{(1-2a)^2}{a^2+(1-a)^2} \geq \frac{23-54a}{25} \iff \frac{25(3a-1)^2(6a+1)}{\geq 0} \geq 0,$ so sum to get the result.

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Cusofay

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Cusofay

#71 h Dec 8, 2023, 11:31 AM

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Set $f(x):= \frac{3-2a}{a^2+(3-a)^2}$ . Due to the homogeneity of the inequality, we can assume $a+b+c=3$ . The inequality can now be rewritten as follows:

$\sum_{cyc} f(a) \geq \frac{3}{5}$
This is true by tangent trick lemma since:
$f(x)\geq f'(1)(x-1)+f(1)$ $\iff \frac{3-2x}{x^2+(3-x)^2} \geq \frac{1}{5}-\frac{18}{25}(x-1)$
$\iff \frac{18(x-1)^2(2x+1)}{25(2x^2-6x+9)} \geq 0$
Which holds for all $0<x<3$

$\mathbb{Q.E.D.}$

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how_to_what_to

61 posts

how_to_what_to

#72 h Dec 11, 2023, 6:28 AM • 1 Y

Y by MihaiT

shaaam wrote:

I think, cauchy-Swarchz inequality can be used to prove this inequality...

Problem.

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

Solution: Let :

$P = \frac {\left(b + c - a\right)^{2}}{\left(b + c\right)^{2} + a^{2}} + \frac {\left(c + a - b\right)^{2}}{\left(c + a\right)^{2} + b^{2}} + \frac {\left(a + b - c\right)^{2}}{\left(a + b\right)^{2} + c^{2}}$
By Cauchy Swarchz inequality, we have

$(a^2 + b^2 + c^2 + (a + b)^2 + (b + c)^2 + (c + a)^2)(P) \geq (a + b + c)^2$

$(3(a^2 + b^2 + c^2) + 2(ab + bc + ca))(P) \geq (a + b + c)^2$

$(3(a + b + c)^2 - 4(ab + bc + ca))(P) \geq (a + b + c)^2$

$P \geq \frac {(a + b + c)^2}{3(a + b + c)^2 - 4(ab + bc + ca)^2}$

but we know that $(a + b + c)^2 \geq 3(ab + bc + ca)$

$P\geq \frac {3(ab + bc + ca)}{9(ab + bc + ca) - 4(ab + bc + ca)} = \frac{3}{5}$

If my solution contains error, can you please correct it? Thanks .

however, you can not turn into $P\geq \frac {3(ab + bc + ca)}{9(ab + bc + ca) - 4(ab + bc + ca)} = \frac{3}{5}$ ,cause it’s on the denominator

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Math4Life7

1703 posts

Math4Life7

#73 h Feb 27, 2024, 2:52 AM

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Homogenous so we let $a+b+c = 3$ . We get $\frac{(b+c-a)^2}{a^2+(b+c)^2} = \frac{(3-2a)^2}{2a^2 - 6a + 9} = 2 - \frac{9}{2a^2-6a+9}$ . Thus we need to prove that $\sum_{\text{cyc}} \frac{1}{2a^2-6a+9} \leq \frac{3}{5}$ We claim that $\frac{1}{2a^2-6a+9} \leq \frac{2a+3}{25}$ Obviously this proves the problem. Multiplying both sides by $50a^2-150a+225$ we get $4a^3-6a^2+2 \geq 0 \Rightarrow 2(x-1)^2(2x+1) \geq 0$ This is obviously true over $[0, 3]$ and we have the result. $\blacksquare$

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joshualiu315

2513 posts

joshualiu315

#75 h Dec 30, 2024, 6:11 AM

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The inequality is homogenous, so we let $a+b+c=3$ . It suffices to prove

$\sum_{\text{cyc}} \frac{(3-2a)^2}{2a^2-6a+9} \ge \frac{3}{5}.$
Suppose that $f(x) = \tfrac{(3-2x)^2}{2x^2-6x+9}$ . Apply the tangent line trick with $a = \tfrac{a+b+c}{3} = 1$ :

$\frac{(3-2x)^2}{2x^2-6x+9} \ge \frac{1}{5} - \frac{18}{25}(x-1),$ $\iff \frac{18(x-1)^2(2x+1)}{25(2x^2-6x+9)} \ge 0,$
which is obviously true for $x \in (0,3)$ . $\blacksquare$

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Maximilian113

524 posts

Maximilian113

#76 h Jan 30, 2025, 4:27 AM • 1 Y

Y by teomihai

The inequality is homogenous, so assume WLOG that $a+b+c=3.$ Then the inequality becomes $\sum \frac{(3-2a)^2}{a^2+(3-a)^2} \geq \frac35.$ However, observe that $\frac{(3-2x)^2}{x^2+(3-x)^2} \geq -\frac{18}{25}x+\frac{23}{25}$ for $x \geq -\frac12,$ this can be proven from direct expansion and then factoring the resulting polynomial. Thus summing this up for $x=a, b, c$ yields the desired result. QED

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teomihai

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teomihai

#77 h Jan 30, 2025, 5:16 AM

Y by

Maximilian113 wrote:

The inequality is homogenous, so assume WLOG that $a+b+c=3.$ Then the inequality becomes $\sum \frac{(3-2a)^2}{a^2+(3-a)^2} \geq \frac35.$ However, observe that $\frac{(3-2x)^2}{x^2+(3-x)^2} \geq -\frac{18}{25}x+\frac{23}{25}$ for $x \geq -\frac12,$ this can be proven from direct expansion and then factoring the resulting polynomial. Thus summing this up for $x=a, b, c$ yields the desired result. QED

Yes ,indeede with TLM.

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Marcus_Zhang

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Marcus_Zhang

#78 h Apr 1, 2025, 1:40 AM • 1 Y

Y by teomihai

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