- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,909

Let $x, y$ be real numbers such that $9x^2+8xy+7y^2 \le 6$.

Show that $7x+12xy+5y \le 9$.

Show that $7x+12xy+5y \le 9$.

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,909

Let $x, y$ be real numbers such that $9x^2+8xy+7y^2 \le 6$.

Show that $7x+12xy+5y \le 9$.

Show that $7x+12xy+5y \le 9$.

- Moderator
- #2

- Feb 7, 2012

- 2,786

[graph]1hbxwofxmo[/graph] [Click on the graph for a larger version.]

The blue curve is the ellipse $9x^2+8xy+7y^2 = 6$. The brown curve is the hyperbola $7x+12xy+5y = 9$. They share a common tangent at the point $(1/2,1/2)$, with gradient $-13/11.$ The interior of the ellipse is the region $9x^2+8xy+7y^2 \leqslant 6$, and the region between the two branches of the hyperbola is given by $7x+12xy+5y \leqslant 9$. From the geometry, it is clear that the first of those regions is contained in the second. I don't see a neat way to prove that algebraically, but I think this is one of those cases where a picture says more than an equation could.

- Jan 25, 2013

- 1,225

Let $x, y$ be real numbers such that $9x^2+8xy+7y^2 \le 6$.

Show that $7x+12xy+5y \le 9$.---(2)

Let $x, y$ be real numbers such that $9x^2+8xy+7y^2 \le 6 ----(1)$.

in fact we only have to discuss :x>0, and y>0

$9x^2+8xy+7y^2 \le 9x^2+4x^2+4y^2+7y^2$ (GM$\le AM)$

the equivalence exists only if x=y

so we have :$24x^2\le 6$

$\therefore x=y\le \dfrac{1}{2}$

$\therefore $ the left side of (2) :$\dfrac{7}{2}+\dfrac{12}{4}+\dfrac{5}{2}\le 9$

- Thread starter
- Admin
- #4

- Feb 14, 2012

- 3,909

Thank you to both of you,Let $x, y$ be real numbers such that $9x^2+8xy+7y^2 \le 6$.

Show that $7x+12xy+5y \le 9$.

I think the solution proposed by other that wanted to share here is very similar to the concept of

If we let $x=a+\dfrac{1}{2}$ and $y=b+\dfrac{1}{2}$, where $a, b \in R$, the given inequality becomes

$9x^2+8xy+7y^2 \le 6$

$9\left( a+\dfrac{1}{2} \right)^2+8\left( a+\dfrac{1}{2} \right)\left( b+\dfrac{1}{2} \right)+7\left( b+\dfrac{1}{2} \right)^2 \le 6$

$13a+11b+8ab \le -(9a^2+7b^2)$

The inequality that we wanted to prove, its LHS after undergo the transformation turns to

$7x+12xy+5y=7\left( a+\dfrac{1}{2} \right)+12\left( a+\dfrac{1}{2} \right)\left( b+\dfrac{1}{2} \right)+5\left( b+\dfrac{1}{2} \right)=13a+11b+8ab+9 \le 9-(9a^2+7b^2) \le 9$

Adding 9 to both sides of the inequality $13a+11b+8ab \le -(9a^2+7b^2)$ gives

$7x+12xy+5y=13a+11b+8ab+9 \le 9-(9a^2+7b^2) \le 9$

and we are done.