Titu's, CS, or Something Else?

Algebra Level 4

The real numbers a , b , c , x , y , z a , b , c , x , y , z satisfy a ≥ b ≥ c > 0 a\geq b \geq c > 0 and x ≥ y ≥ z > 0 x \geq y \geq z >0

The minimum value of

( a x ) 2 ( b y + c z ) ( b z + c y ) + ( b y ) 2 ( c z + a x ) ( c x + a z ) + ( c z ) 2 ( a x + b y ) ( a y + b x ) {\dfrac{(ax)^{2}}{(by + cz)(bz + cy)}} + {\dfrac{(by)^{2}}{(cz + ax)(cx + az)}} + {\dfrac{(cz)^{2}}{(ax + by)(ay + bx)}}

can be expressed as m n \frac{m}{n} , where m m and n n are positive coprime integers. Find 50 ( m + n ) 50(m + n)


The answer is 350.

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U Z by U Z , 24, Guyana

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3 solutions

Kartik Sharma
Dec 12, 2014

( a x ) 2 ( b y + c z ) ( b z + c y ) + ( b y ) 2 ( a x + c z ) ( c x + a z ) + ( c z ) 2 ( a x + b y ) ( a y + b x ) \frac{{(ax)}^{2}}{(by+cz)(bz+cy)} + \frac{{(by)}^{2}}{(ax+cz)(cx+az)} + \frac{{(cz)}^{2}}{(ax+by)(ay+bx)}

First thing that came in my mind was Cauchy Schwarz,

( ( a x ) 2 ( b y + c z ) ( b z + c y ) + ( b y ) 2 ( a x + c z ) ( c x + a z ) + ( c z ) 2 ( a x + b y ) ( a y + b x ) ) ( ( b y + c z ) ( b z + c y ) ( a x ) 2 + ( a x + c z ) ( c x + a z ) ( b y ) 2 + ( a x + b y ) ( a y + b x ) ( c z ) 2 ) ≥ 9 (\frac{{(ax)}^{2}}{(by+cz)(bz+cy)} + \frac{{(by)}^{2}}{(ax+cz)(cx+az)} + \frac{{(cz)}^{2}}{(ax+by)(ay+bx)})(\frac{(by+cz)(bz+cy)}{{(ax)}^{2}} + \frac{(ax+cz)(cx+az)}{{(by)}^{2}} + \frac{(ax+by)(ay+bx)}{{(cz)}^{2}}) \geq 9

As x ≥ y ≥ z > 0 x \geq y \geq z >0 and a ≥ b ≥ c > 0 a \geq b \geq c > 0 ,

( b y + c z ) ≥ ( b z + c y ) (by+cz) \geq (bz + cy) and so on.

Hence, the above equation becomes

( ( a x ) 2 ( b y + c z ) ( b z + c y ) + ( b y ) 2 ( a x + c z ) ( c x + a z ) + ( c z ) 2 ( a x + b y ) ( a y + b x ) ) ( ( b z + c y ) 2 ( a x ) 2 + ( c x + a z ) 2 ( b y ) 2 + ( a y + b x ) 2 ( c z ) 2 ) ≥ 9 (\frac{{(ax)}^{2}}{(by+cz)(bz+cy)} + \frac{{(by)}^{2}}{(ax+cz)(cx+az)} + \frac{{(cz)}^{2}}{(ax+by)(ay+bx)})(\frac{{(bz+cy)}^{2}}{{(ax)}^{2}} + \frac{{(cx + az)}^{2}}{{(by)}^{2}} + \frac{{(ay+bx)}^{2}}{{(cz)}^{2}}) \geq 9

Now, using AM-GM on the 2nd productend,

( b z + c y ) 2 ≥ 4 b y c z {(bz+cy)}^{2} \geq 4bycz and so on.

Hence, the above equation simplifies to

( ( a x ) 2 ( b y + c z ) ( b z + c y ) + ( b y ) 2 ( a x + c z ) ( c x + a z ) + ( c z ) 2 ( a x + b y ) ( a y + b x ) ) ( 4 b y c z ( a x ) 2 + 4 c x a z ( b y ) 2 + 4 a x b y ( c z ) 2 ) ≥ 9 (\frac{{(ax)}^{2}}{(by+cz)(bz+cy)} + \frac{{(by)}^{2}}{(ax+cz)(cx+az)} + \frac{{(cz)}^{2}}{(ax+by)(ay+bx)})(\frac{4bycz}{{(ax)}^{2}} + \frac{4cxaz}{{(by)}^{2}} + \frac{4axby}{{(cz)}^{2}}) \geq 9

Now again using AM-GM,

b y c z ( a x ) 2 + c x a z ( b y ) 2 + a x b y ( c z ) 2 ≥ 3 a 2 b 2 c 2 x 2 y 2 z 2 a 2 b 2 c 2 x 2 y 2 z 2 3 \frac{bycz}{{(ax)}^{2}} + \frac{cxaz}{{(by)}^{2}} + \frac{axby}{{(cz)}^{2}} \geq 3\sqrt [ 3 ]{ \frac { { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }{ x }^{ 2 }{ y }^{ 2 }{ z }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }{ x }^{ 2 }{ y }^{ 2 }{ z }^{ 2 } } }

≥ 3 \geq 3

As a result equation simplifies

( ( a x ) 2 ( b y + c z ) ( b z + c y ) + ( b y ) 2 ( a x + c z ) ( c x + a z ) + ( c z ) 2 ( a x + b y ) ( a y + b x ) ) ( 12 ) ≥ 9 (\frac{{(ax)}^{2}}{(by+cz)(bz+cy)} + \frac{{(by)}^{2}}{(ax+cz)(cx+az)} + \frac{{(cz)}^{2}}{(ax+by)(ay+bx)})(12) \geq 9

( ( a x ) 2 ( b y + c z ) ( b z + c y ) + ( b y ) 2 ( a x + c z ) ( c x + a z ) + ( c z ) 2 ( a x + b y ) ( a y + b x ) ) ≥ 3 4 (\frac{{(ax)}^{2}}{(by+cz)(bz+cy)} + \frac{{(by)}^{2}}{(ax+cz)(cx+az)} + \frac{{(cz)}^{2}}{(ax+by)(ay+bx)}) \geq \frac{3}{4}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Yeah! Kartik sharma

U Z - 6 years, 6 months ago

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Well, nice problem!

Kartik Sharma - 6 years, 6 months ago

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Excellent solution, Kartik Sharma.

Saurabh Mallik - 5 years, 3 months ago
Pranjal Jain
Dec 11, 2014

I just substituted a=b=c and x=y=z and got the value as 3 4 \frac{3}{4}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you please explain why ? Thanks !

Deepanshu Gupta - 6 years, 6 months ago

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because most inequalities work that way!

Agnishom Chattopadhyay - 6 years, 6 months ago

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Yes! You are right! Lol!

Pranjal Jain - 6 years, 6 months ago

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@Pranjal Jain – Please think its a good question @Agnishom Chattopadhyay @Pranjal Jain

U Z - 6 years, 6 months ago

Correct me if I am wrong, but I think this only works because x is greater or equal than y and so on. If x would have been just greater I don't think that have worked.

Radinoiu Damian - 6 years, 5 months ago

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If you live in Romania, you definitely live on Earth!! I mean that if something is true for even an "equality" case, it must be true for "greater" case.

Pranjal Jain - 6 years, 5 months ago
Barack Clinton
Feb 1, 2015

Cauchy-Schwarz ⟹ ( b y + c z ) ≤ ( b 2 + c 2 ) ( y 2 + z 2 ) \implies (by+cz)\le\sqrt{(b^2+c^2)(y^2+z^2)}

and ( b y + c z ) ≤ ( b 2 + c 2 ) ( y 2 + z 2 ) (by+cz)\le\sqrt{(b^2+c^2)(y^2+z^2)} , do the same with the other denominators.

Using Chebyshev inequality we maximize again the denominator;

( b 2 + c 2 ) ( y 2 + z 2 ) 2 ≤ ( b 2 y 2 + c 2 z 2 ) = ( b y ) 2 + ( c y ) 2 \frac{(b^2+c^2)(y^2+z^2)}{2}\le(b^2y^2+c^2z^2)=(by)^2+(cy)^2

Now we obtained:

( a x ) 2 2 ( ( b y ) 2 + ( c y ) 2 ) ) + ( b y ) 2 2 ( ( a x ) 2 + ( c z ) 2 ) ) + ( c z ) 2 2 ( ( a x ) 2 + ( b y ) 2 ) ) \frac{(ax)^2}{2\left((by)^2+(cy)^2)\right)}+\frac{(by)^2}{2\left((ax)^2+(cz)^2)\right)}+\frac{(cz)^2}{2\left((ax)^2+(by)^2)\right)}

applying Nesbitt's Inequality (can be found here ) the sum above is greater or equal to 3 4 \frac{3}{4}

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