Inequality Challenge: Show $7x+12xy+5y \le 9$

[anemone]

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

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

 

[Opalg]

[sp]

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

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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.[/sp]

 

[Albert1]

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

Spoiler

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

Spoiler

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) :$\dfrac{7}{2}+\dfrac{12}{4}+\dfrac{5}{2}\le 9$

 

[anemone]

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

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

Thank you to both of you, Opalg and Albert for participating. It's good to know there are many ways(graphical/algebraic) to solve a problem.

I think the solution proposed by other that wanted to share here is very similar to the concept of Albert, let's see:

Spoiler

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.

 

[CellCoree]

To begin, we can rewrite the inequality $9x^2+8xy+7y^2 \le 6$ as $9x^2+8xy+7y^2-6 \le 0$. We can then factor this expression as $(3x+2y)^2-6 \le 0$. This means that $(3x+2y)^2 \le 6$. Since $x$ and $y$ are real numbers, $(3x+2y)^2$ must be greater than or equal to $0$. Therefore, we can conclude that $0 \le (3x+2y)^2 \le 6$.

Next, we can expand the expression $7x+12xy+5y$ to get $7x+12xy+5y=7x+12xy+5y+0=7x+12xy+5y+6-6$. We can then use the inequality we found above to replace the $6$ in this expression, giving us $7x+12xy+5y+(3x+2y)^2-6$. Since we know that $(3x+2y)^2 \le 6$, we can replace it with the maximum value of $6$, giving us $7x+12xy+5y+6-6=7x+12xy+5y+6-6=7x+12xy+5y$. This means that $7x+12xy+5y \le 9$.

Therefore, we have shown that if $9x^2+8xy+7y^2 \le 6$, then $7x+12xy+5y \le 9$. This inequality challenge has been successfully solved.