Let $x$ and $y$ be positive real numbers with $x + y =1 $. Prove that $$\frac{(3x-1)^2}{x}+ \frac{(3y-1)^2}{y} \ge1.$$For which $x$ and $y$ equality holds? (K. Czakler, GRG 21, Vienna)
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sqing
03.10.2021 12:51
parmenides51 wrote: Let $x$ and $y$ be positive real numbers with $x + y =1 $. Prove that $$\frac{(3x-1)^2}{x}+ \frac{(3y-1)^2}{y} \ge1.$$For which $x$ and $y$ equality holds? (K. Czakler, GRG 21, Vienna) ALBANIA - 2015 https://artofproblemsolving.com/community/c6h1664258p16611930 $$\frac{(3a-1)^2}{a}+\frac{(3b-1)^2}{b}\geq \frac{(|3a-1|+|3b-1|)^2}{a+b}\geq (3a-1+3b-1)^2=1$$ Let $x,y>0,x+y=1$ .Prove that $$\frac{(3x - 1)^2}{y}+\frac{(3y-1)^2}{x}\geq 1$$$$\frac{(3x - 1)^2}{2y}+\frac{(3y-1)^2}{x}>\frac{1}{2}$$
Manolis
03.10.2021 13:12
sqing wrote: parmenides51 wrote: Let $x$ and $y$ be positive real numbers with $x + y =1 $. Prove that $$\frac{(3x-1)^2}{x}+ \frac{(3y-1)^2}{y} \ge1.$$For which $x$ and $y$ equality holds? (K. Czakler, GRG 21, Vienna) ALBANIA - 2015 https://artofproblemsolving.com/community/c6h1664258p16611930 $$\frac{(3a-1)^2}{a}+\frac{(3b-1)^2}{b}\geq \frac{(|3a-1|+|3b-1|)^2}{a+b}\geq (3a-1+3b-1)^2=1$$ Let $x,y>0,x+y=1$ .Prove that $$\frac{(3x - 1)^2}{y}+\frac{(3y-1)^2}{x}\geq 1$$$$\frac{(3x - 1)^2}{2y}+\frac{(3y-1)^2}{x}>\frac{1}{2}$$ $\frac{(3x-1)^2}{y}+\frac{(3y-1)^2}{x}\geq \frac{(|3x-1|+|3y-1|)^2}{y+x}\geq (3x-1+3y-1)^2=1$ $\frac{(3x-1)^2}{2y}+\frac{(3y-1)^2}{x}\geq \frac{(|3x-1|+|3y-1|)^2}{2y+x}> \frac{(|3x-1|+|3y-1|)^2}{2y+2x}\geq \frac{(3x-1+3y-1)^2}{2}=\dfrac{1}{2}$
sqing
03.10.2021 13:39
Manolis wrote:
sqing wrote:
parmenides51 wrote:
Let $x$ and $y$ be positive real numbers with $x + y =1 $. Prove that
$$\frac{(3x-1)^2}{x}+ \frac{(3y-1)^2}{y} \ge1.$$For which $x$ and $y$ equality holds?
(K. Czakler, GRG 21, Vienna)
ALBANIA - 2015
https://artofproblemsolving.com/community/c6h1664258p16611930
$$\frac{(3a-1)^2}{a}+\frac{(3b-1)^2}{b}\geq \frac{(|3a-1|+|3b-1|)^2}{a+b}\geq (3a-1+3b-1)^2=1$$
Let $x,y>0,x+y=1$ .Prove that
$$\frac{(3x - 1)^2}{y}+\frac{(3y-1)^2}{x}\geq 1$$$$\frac{(3x - 1)^2}{2y}+\frac{(3y-1)^2}{x}>\frac{1}{2}$$
$\frac{(3x-1)^2}{y}+\frac{(3y-1)^2}{x}\geq \frac{(|3x-1|+|3y-1|)^2}{y+x}\geq (3x-1+3y-1)^2=1$
$\frac{(3x-1)^2}{2y}+\frac{(3y-1)^2}{x}\geq \frac{(|3x-1|+|3y-1|)^2}{2y+x}> \frac{(|3x-1|+|3y-1|)^2}{2y+2x}\geq \frac{(3x-1+3y-1)^2}{2}=\dfrac{1}{2}$
sqing
04.10.2021 06:08
Let $x,y>0,x+y=4$ .Prove that $$\frac{(2x+1)^2}{x}+\frac{(2y+1)^2}{y}\geq 25$$
jasperE3
04.10.2021 06:39
These can also be solved by Jensen.
sqing
04.10.2021 17:49
Let $x,y,z$ be positive real numbers with $x + y+z =1 $. Prove that $$ \frac{(2x-1)^2}{x}+ \frac{(2y-1)^2}{y} + \frac{(2z-1)^2}{z} \ge 1$$$$\frac{(2x-1)^2}{y}+ \frac{(2y-1)^2}{z} + \frac{(2z-1)^2}{x} \ge 1$$
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ab74
10.02.2024 09:25
parmenides51 wrote: Let $x$ and $y$ be positive real numbers with $x + y =1 $. Prove that $$\frac{(3x-1)^2}{x}+ \frac{(3y-1)^2}{y} \ge1.$$For which $x$ and $y$ equality holds? (K. Czakler, GRG 21, Vienna)
Lets work with the left hand side. By creating a common denominator and some expanding in the numerator we get
$$\frac{9x^2y - 6xy + y +9xy^2 - 6xy + x}{xy}$$$$\frac{9xy(x+y)-12xy+1}{xy}$$$$\frac{-3xy+1}{xy}$$(1)
Now, AM GM inequality with $x$ and $y$ gives
$$\frac{x+y}{2} >= \sqrt{xy}$$$$\frac{1}{4} >= xy$$
The maximum value of xy is $\frac{1}{4}$. Plugging this into equation 1 gives 1. The expression can be simplified to $-3 + \frac{1}{xy}$. As xy decreases, the value of this function increases. Thus, the inequality holds.
Equality holds when $x=y=\frac{1}{2}$
sqing
10.02.2024 12:55
Let $ x , y >0 $ and $x + y =1 $. Prove that $$\frac{(3x-1)^2}{x}+ \frac{(3y-1)^2}{y} +xy\geq \frac{5}{4}$$
LLL2019
10.02.2024 13:15
Let $ x , y >0 $ and $x + y =1 $. Prove that $$\frac{(3x-1)^2}{x}+ \frac{(3y-1)^2}{y} + kxy\geq \frac{k^2+4}{4}$$ https://artofproblemsolving.com/community/u488422h3253461p29869075
sqing
10.02.2024 15:13
LLL2019 wrote:
Let $ x , y >0 $ and $x + y =1 $. Prove that
$$\frac{(3x-1)^2}{x}+ \frac{(3y-1)^2}{y} + kxy\geq \frac{k^2+4}{4}$$
https://artofproblemsolving.com/community/u488422h3253461p29869075
sqing
10.02.2024 15:13
Let $ x , y >0 $ and $x + y =1 $. Prove that $$\frac{(2x-3)^2}{x}+ \frac{(2y-3)^2}{y} +144xy\geq 52$$
rhphy_3031
10.02.2024 18:31
sqing wrote: Let $ x , y >0 $ and $x + y =1 $. Prove that $$\frac{(2x-3)^2}{x}+ \frac{(2y-3)^2}{y} +144xy\geq 52$$ $$LHS-RHS=\frac{9(2x-1)^4}{x(1-x)}\ge0$$
sqing
11.02.2024 05:04
rhphy_3031 wrote:
sqing wrote:
Let $ x , y >0 $ and $x + y =1 $. Prove that
$$\frac{(2x-3)^2}{x}+ \frac{(2y-3)^2}{y} +144xy\geq 52$$
$$LHS-RHS=\frac{9(2x-1)^4}{x(1-x)}\ge0$$
DAVROS
11.02.2024 10:57
sqing wrote: Let $ x , y >0 $ and $x + y =1 $. Prove that $\frac{(2x-3)^2}{x}+ \frac{(2y-3)^2}{y} +144xy\geq 52$
$\text{LHS} = \frac{4xy(x+y)+9(x+y)-24xy}{xy}+144xy = \frac9{xy}+144xy -20\ge52$ at $xy=\frac14, x+y=1$
sqing
11.02.2024 17:28
DAVROS wrote:
sqing wrote:
Let $ x , y >0 $ and $x + y =1 $. Prove that $\frac{(2x-3)^2}{x}+ \frac{(2y-3)^2}{y} +144xy\geq 52$
solution$\text{LHS} = \frac{4xy(x+y)+9(x+y)-24xy}{xy}+144xy = \frac9{xy}+144xy -20\ge52$ at $xy=\frac14, x+y=1$
DensSv
29.08.2024 06:04
By Bergstroms inequality we obtain that $\frac{(3x-1)^{2}}{x} +\frac{(3y-1)^{2}}{y} \geq \frac{(3x-1+3y-1)^{2}}{x+y}=1$. The equality case is obtained when x=y, since x+y=1 this means that $x=y=\frac{1}{2}$
Sadigly
01.09.2024 22:05
$\frac{(3x-1)^2}{x}+\frac{(3y-1)^2}{y}\geq\frac{(3x+3y-2)^2}{x+y}=1$