AM-GM test (CNNer in 2015)

Algebra Level 5

If $x$ and $y$ are positive reals such that $x + y \ge 3$ , then find the minimum value of $2x^2 + y^2 + \frac{28}x + \frac1y$ .

The answer is 24.

1 solution

Son Nguyen
Oct 19, 2015

All right We predict that x=2,y=1 If equal sign occurs According AM-GM,we have $y^2+1\geq 2y$ $2x^2+8\geq 8x$

$\frac{1}{y}+y\geq 2$ $\frac{28}{x}+7x\geq 28$ ============> $2x^2+y^2+\frac{28}{x}+\frac{1}{y}+7x+y+9\geq 8x+2y+28+2$ Finally: $2x^2+y^2+\frac{28}{x}+\frac{1}{y}\geq x+y+21\geq\: 24$

Always remember to check the conditions under which you can apply Arithmetic Mean - Geometric Mean . In particular, you need to verify that both terms are positive.

It seems that you are currently missing a condition.

Calvin Lin Staff - 5 years, 7 months ago

Nicely done! I've got a direct solution but it's not as clever as yours

$2{ x }^{ 2 }+\frac { 28 }{ x } +{ y }^{ 2 }+\frac { 1 }{ y } \\ =\left( { x }^{ 2 }+{ x }^{ 2 }+\frac { 8 }{ x } \right) +\left( \frac { { y }^{ 2 } }{ 2 } +\frac { { y }^{ 2 } }{ 2 } +\frac { 1 }{ 2y } \right) +\frac { 20 }{ x } +\frac { 1 }{ 2y } \ge 6x+\frac { 3 }{ 2 } y+\frac { 20 }{ x } +\frac { 1 }{ 2y } =\\ =\left( 5x+\frac { 20 }{ x } \right) +\left( \frac { y }{ 2 } +\frac { 1 }{ 2y } \right) +(x+y)\ge 20+1+3=24$

trongnhan khong - 5 years, 7 months ago

Nice solution

Son Nguyen - 5 years, 7 months ago

Thanks a lot for the beautiful solution!

Adarsh Kumar - 5 years, 7 months ago

you are welcome

Son Nguyen - 5 years, 7 months ago

I think the RHS of second last expression should be $8x + 2y + 30$ . You forgot adding 2.

Anyway, nice solution.

Priyanshu Mishra - 5 years, 7 months ago

thanks.I will repair

Son Nguyen - 5 years, 7 months ago

AM-GM test (CNNer in 2015)

The answer is 24.

1 solution

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