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Algebra Level 4

A = 12 x 3 y + 108 x y 3 + 81 y 4 + x 4 + 1 + 2 y + y 2 + 54 x 2 y 2 x 2 + x 2 y + 6 x y + 9 y 3 + 6 x y 2 + 9 y 2 A = \dfrac{12 x^{3} y + 108x y^{3} + 81y^{4} + x^{4} + 1 + 2y + y^{2} + 54x^{2} y^{2} }{x^{2} + x^{2} y + 6xy + 9y^{3} + 6xy^{2} + 9y^{2}}

Let T T be the minimum value of A A for positive real numbers x , y x,y . Find the value of 5 T 5T .


The answer is 10.

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Harsh Shrivastava by Harsh Shrivastava , 20, India

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6 solutions

Tanishq Varshney
May 11, 2015

A = ( x + 3 y ) 4 + ( y + 1 ) 2 x 2 ( y + 1 ) + 6 x y ( y + 1 ) + 9 y 2 ( y + 1 ) \huge{ A = \frac{(x+3y)^4+(y+1)^2}{x^2(y+1)+6xy(y+1)+9y^2(y+1)}}

= ( x + 3 y ) 4 + ( y + 1 ) 2 ( x + 3 y ) 2 ( y + 1 ) \large{\frac{(x+3y)^4+(y+1)^2}{(x+3y)^2(y+1)}}

= ( x + 3 y ) 2 ( y + 1 ) + ( y + 1 ) ( x + 3 y ) 2 \large{\frac{(x+3y)^2}{(y+1)} +\frac{ (y+1)}{(x+3y)^2}}

Applying A M − G M AM-GM

( x + 3 y ) 2 ( y + 1 ) + ( y + 1 ) ( x + 3 y ) 2 2 ≥ 1 \large{\frac{\frac{(x+3y)^2}{(y+1)} +\frac{ (y+1)}{(x+3y)^2}}{2}\geq \sqrt{1}}

T ≥ 2 T\geq 2

13 Helpful 0 Interesting 0 Brilliant 0 Confused

First of all, you have a typo in your solution. It should be,

A ≥ 2 ⟹ T = min ⁡ ( A ) = 2 \color{#D61F06}{A}\geq 2\implies T=\min(A)=2

Second of all, since we have to find minimum, you need to show that the equality case does exist for some x , y ∈ R + x,y\in\Bbb{R^+} as required in the problem. Otherwise, your solution is incomplete. Showing the equality case is easy though since you just use the equality case for AM-GM to show it. Interesting enough, there isn't just one but multiple equality cases for the problem (you can see that by graphing the equality case equation). Here's a part of the equality case graph which is relevant to our problem:

\qquad\qquad\qquad\qquad Image Image

This is the part of the curve ( x + 3 y ) 4 = ( y + 1 ) 2 (x+3y)^4=(y+1)^2 for positive real x , y x,y . (I admit that the image quality and accuracy of the graph is quite bad, but you can get the idea). Note that you cannot use the equality case x = 1 , y = 0 x=1,y=0 since 0 ∉ R + 0\notin\Bbb{R^+} .

Prasun Biswas - 6 years ago
Adarsh Kumar
May 11, 2015

In the numerator,after seeing the terms 81 y 4 , x 4 81y^4,x^4 ,we get that the numerator has some relation to ( x + 3 y ) 4 (x+3y)^4 ,which when we expand get that all the terms in the numerator apart from y 2 , 2 y , 1 y^2,2y,1 have been accounted for.Hence the numerator becomes: ( x + 3 y ) 4 + y 2 + 2 y + 1 = ( x + 3 y ) 4 + ( y + 1 ) 2 (x+3y)^4+y^2+2y+1\\ =(x+3y)^4+(y+1)^2 .Now,moving on to the denominator,first thing we notice is that it has the terms, ( 3 y ) 2 , x 2 , 6 x y (3y)^2,x^2,6xy ,which are present in the expansion of ( x + 3 y ) 2 (x+3y)^2 .Hence,the denominator becomes, ( x + 3 y ) 2 + x 2 y + 9 y 3 + 6 x y 2 = ( x + 3 y ) 2 + y ( x 2 + 9 y 2 + 6 x y ) = ( x + 3 y ) 2 + y ( ( x + 3 y ) 2 ) = ( x + 3 y ) 2 ( y + 1 ) (x+3y)^2+x^2y+9y^3+6xy^2\\ =(x+3y)^2+y(x^2+9y^2+6xy)\\ =(x+3y)^2+y((x+3y)^2)\\ =(x+3y)^2(y+1) .Now,the fraction has been converted to, ( x + 3 y ) 4 + ( y + 1 ) 2 ( x + 3 y ) 2 ( y + 1 ) = ( x + 3 y ) 2 ( y + 1 ) + ( y + 1 ) ( x + 3 y ) 2 \dfrac{(x+3y)^4+(y+1)^2}{(x+3y)^2(y+1)}\\ =\dfrac{(x+3y)^2}{(y+1)}+\dfrac{(y+1)}{(x+3y)^2} .Now,one can simply apply, A.M-G.M \text{A.M-G.M} .

6 Helpful 0 Interesting 0 Brilliant 0 Confused

to get idea on factoring polynomials easily
read this

Abhinav Raichur - 6 years, 1 month ago

A = ( x + 3 y ) 4 + ( y + 1 ) 2 x 2 ( y + 1 ) + 6 x y ( y + 1 ) + 9 y 2 ( y + 1 ) \large{ A = \dfrac{(x+3y)^4+(y+1)^2}{x^2(y+1)+6xy(y+1)+9y^2(y+1)}}

= ( x + 3 y ) 4 + ( y + 1 ) 2 ( x + 3 y ) 2 ( y + 1 ) \large{\dfrac{(x+3y)^4+(y+1)^2}{(x+3y)^2(y+1)}}

= ( x + 3 y ) 2 ( y + 1 ) + ( y + 1 ) ( x + 3 y ) 2 \large{\dfrac{(x+3y)^2}{(y+1)} +\dfrac{ (y+1)}{(x+3y)^2}} M i n i m u m o f 1 x + x f o r x > 0 i s a l w a y s = 2. Minimum ~of~ \dfrac 1 x + x ~ for~ x>0~ is~ always =2.\\

∴ 5 T = 5 ∗ 2 = 10 \therefore 5T=5*2= \Huge \color{#EC7300}{10}

f ( x ) = 1 x + x , f ′ ( x ) = − 1 x 2 + 1 = 0. ∴ x = ± 1. f " ( x ) = 1 x 3 > 0 i f x > 0 , a n d f " ( x ) < 0 i f x < 0. ∴ x = 1 i s m i n i m u m a n d f ( 1 ) m i n = 1 x + x = 2. ∴ x = − 1 i s m a x i m u m a n d f ( − 1 ) m a x = 1 x + x = − 2. f(x)= \dfrac 1 x + x ~ ,~~~~~~~~f'(x)=- \dfrac 1 {x^2} + 1=0.~~~~~~~\therefore~x=\pm 1.\\f"(x)=\dfrac 1 {x^3}>0~~if ~~x>0,~~~~and~~~~f"(x)<0~~if~~x<0.\\\therefore~x=1~ is~minimum~~~~~and~~~\color{#3D99F6}{ f(1)_{min}= \dfrac 1 x + x =2.}\\\therefore~x= - 1~ is~maximum~~~~~and~~~\color{#D61F06}{ f(-1)_{max}= \dfrac 1 x + x = - 2.}
E v e n a l g i b r i c a l l y , f o r Δ > 0 i f x > 1 s a y x = 1 + Δ , x + 1 x = 1 + 2 ∗ Δ + Δ 2 + 1 1 + Δ = 2 Δ + 2 + Δ 2 1 + Δ > 2 S i m i l a r l y f o r Δ < 0 i f x < − 1 , x + 1 x < − 2. Even~~algibrically,~~for~~\Delta>0 ~~ if~~\color{#3D99F6}{ x>1} ~~say~x=1+\Delta,\\\color{#3D99F6}{ x+\dfrac 1 x} =\dfrac{1+2*\Delta+\Delta^2+1}{1+\Delta}=\dfrac{2\Delta+2+\Delta^2}{1+\Delta}\color{#3D99F6}{ >2}\\Similarly~for~\Delta<0~~if~~\color{#D61F06}{x< - 1,~~~~~~~x+\dfrac 1 x< - 2.}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Akshay Sharma
Dec 23, 2015

This is really a unique question.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Mini Gupta
May 11, 2015

Was just listening to the song. ..!!!! This moment. ..We own it. ..

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Li Yuelin
May 11, 2015

By simplification, A = [(x+3y)^4+(y+1)^4]/[x^2(y+1)+6xy(y+1)+9y^2(y+1)] = [(x+3y)^4+(y+1)^2]/[(x+3y)^2(y+1)] = (x+3y)^2/(y+1) + (y+1)/(x+3y)^2 By AM-GM inequality, T = 2 Thus 5T = 10

0 Helpful 0 Interesting 0 Brilliant 0 Confused

A = ( x + 3 y ) 4 + ( y + 1 ) 2 x 2 ( y + 1 ) + 6 x y ( y + 1 ) + 9 y 2 ( y + 1 ) \huge{ A = \frac{(x+3y)^4+(y+1)^2}{x^2(y+1)+6xy(y+1)+9y^2(y+1)}}

= ( x + 3 y ) 4 + ( y + 1 ) 2 ( x + 3 y ) 2 ( y + 1 ) \large{\frac{(x+3y)^4+(y+1)^2}{(x+3y)^2(y+1)}}

= ( x + 3 y ) 2 ( y + 1 ) + ( y + 1 ) ( x + 3 y ) 2 \large{\frac{(x+3y)^2}{(y+1)} +\frac{ (y+1)}{(x+3y)^2}}

Applying A M − G M AM-GM

T = 2 T=2

Tanishq Varshney - 6 years, 1 month ago

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That's nice, post it as solution! @Tanishq Varshney

Harsh Shrivastava - 6 years, 1 month ago

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Ok i'll post it

Tanishq Varshney - 6 years, 1 month ago

Did exact same...

Rushikesh Joshi - 6 years ago

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