MHB Inequality Involves The Sides Of Triangle

[anemone]

Let $a,\,b$ and $c$ be the sides of a triangle and $x,\,y$ and $z$ are real numbers such that $x+ y+ z = 0$. Prove that $a^2yz +b^2xz+c^2xy\le 0$.

 

[Euge]

Here is my solution.

Spoiler

Without loss of generality, assume $c$ is the longest side of the triangle. If the triangle is obtuse or right, then $c^2 \ge a^2 + b^2$. So since $z = -x - y$, then $a^2yz + b^2xz + c^2xy = -(a^2y^2 - b^2x^2) + (c - a^2 - b^2)xy.$ Now $a^2y^2 + b^2x^2 \ge 2ab|xy| = [(a + b)^2 - a^2 - b^2]|xy| \ge (c^2 - a^2 - b^2)|xy|,$ where the last step follows from the triangle inequality $a + b > c$. Thus $-(a^2y^2 + b^2x^2) \le -(c^2 - a^2 - b^2)|xy|$, forcing

$$-(a^2y^2 + b^2x^2) + (c^2 - a^2 - b^2)xy \le -(c^2 - a^2 - b^2)|xy| + (c^2 - a^2 - b^2)xy = (c^2 - a^2 - b^2)[-|xy| + xy] \le 0.$$

Now suppose the triangle is acute. Then $c^2 < a^2 + b^2$. Since $c$ is the longest side, we also have $b^2 < a^2 + c^2$, and $a^2 < b^2 + c^2$. We can write

$$xy = \frac{z^2 - x^2 - y^2}{2},\quad yz = \frac{x^2 - y^2 - z^2}{2},\quad \text{and} \quad xz = \frac{y^2 - x^2 - z^2}{2},$$

which are derived from the equations $z^2 = (x + y)^2$, $x^2 = (y + z)^2$, and $y^2 = (x + z)^2$, respectively. Thus

$$a^2yz + b^2xz + c^2xy = -\frac{(b^2 + c^2 - a^2)x^2 + (c^2 + a^2 - b^2)y^2 + (a^2 + b^2 - c^2)z^2}{2} \le 0.\quad \text{QED}$$

 

[anemone]

Well done Euge! And thanks for participating!

Here's my solution:

Spoiler

Since $a,\,b$ and $c$ are the lengths of a triangle we have:

$a^2≤(b+c)^2, \,b^2≤(a+c)^2, \,c^2≤(a+b)^2$ and so

$\begin{align*}a^2x+b^2y+c^2z&\le (b+c)^2x+(a+c)^2y+(a+b)^2z\\&=a^2(y+z)+b^2(x+z)+c^2(x+y)+2abz+2acy+2bcx\\&=a^2(-x)+b^2(-y)+c^2(-z)+2abz+2acy+2bcx\end{align*}$

and so our mission now has turned into proving

$2(a^2x+b^2y+c^2z)\le 2abz+2acy+2bcx$, i.e.

$a^2x+b^2y+c^2z \le abz+acy+bcx$, but that is true from the Chebyshev's inequality which says if

$ab\ge ac \ge bc$ and $x\ge y \ge z$,

$$abz+acy+bcx \le \frac{(ab+ac+bc)(x+y+z)}{3}=0$$ since $x+y+z=0.$

Therefore we have:

$a^2x+b^2y+c^2z \le abz+acy+bcx \le 0$ (Q.E.D.)

 

[lfdahl]

Well done Euge! And thanks for participating!

Here's my solution:

Spoiler

Since $a,\,b$ and $c$ are the lengths of a triangle we have:

$a^2≤(b+c)^2, \,b^2≤(a+c)^2, \,c^2≤(a+b)^2$ and so

$\begin{align*}a^2x+b^2y+c^2z&\le (b+c)^2x+(a+c)^2y+(a+b)^2z\\&=a^2(y+z)+b^2(x+z)+c^2(x+y)+2abz+2acy+2bcx\\&=a^2(-x)+b^2(-y)+c^2(-z)+2abz+2acy+2bcx\end{align*}$

and so our mission now has turned into proving

$2(a^2x+b^2y+c^2z)\le 2abz+2acy+2bcx$, i.e.

$a^2x+b^2y+c^2z \le abz+acy+bcx$, but that is true from the Chebyshev's inequality which says if

$ab\ge ac \ge bc$ and $x\ge y \ge z$,

$$abz+acy+bcx \le \frac{(ab+ac+bc)(x+y+z)}{3}=0$$ since $x+y+z=0.$

Therefore we have:

$a^2x+b^2y+c^2z \le abz+acy+bcx \le 0$ (Q.E.D.)

Thankyou very much anemone and Euge

Spoiler

The purpose of the challenge was to prove, that $a^2yz+b^2zx+c^2xy \le 0$ but it seems to me, that anemones end result is a different one?? Maybe I am wrong ...? Thankyou very much for clearing the matter.

 

[Euge]

Hi lfdahl,

Spoiler

It appears anemone tried to prove a different inequality than the one in the given problem, but even so there is an error in that solution: since $x$, $y$, and $z$ are arbitrary real numbers, it is not necessarily true that $a^2x + b^2y + c^2z \le (b + c)^2x + (c + a)^2y + (a + b)^2z$. In fact, if $x$, $y$, and $z$ are negative, then $a^2x + b^2y + c^2z > (b + c)^2x + (c + a)^2y + (a + b)^2z$. Take for instance $a = 3$, $b = 4$, $c = 5$, and $x = y = z = -1$. Then $a^2x + b^2y + c^2z = -50$, but $(b + c)^2x + (c + a)^2y + (a + b)^2z = -155 < -50$.

 

[lfdahl]

Thank you Euge, but

Spoiler

$x = y = z = -1$ is not allowed, albeit: $x + y + z = 0$

 

[Euge]

Oops, I accidentally edited your post with my message, rather than post mine with a "reply with quote"! Ifdahl, if you could please, rewrite your statement over mine in an edit.

Edit: I tried to put back your question -- does it look correct? My response was this:

Spoiler

Sorry, I overlooked the condition $x + y + z = 0$. But the result will not hold: If $a = 3$, $b = 4$, $c = 5$, $x = -1$, $y = 1$, and $z = 0$, then $x + y + z = 0$ and the triangle with sides $a,b,c$ form a right triangle, but $a^2x + b^2y + c^2z = -9 + 16 + 0 = 7 > 0$.

 

[lfdahl]

Thankyou Euge for the edit!My suggested solution:

Spoiler

WLOG, I can choose $x,y$ and $z$ such, that: $yz \ge zx \ge xy$. On the other hand, I can also choose the sides in a triangle, such that: $a^2 \le b^2 \le c^2$.

Applying Chebyshev´s sum inequality:\[3\left ( a^2yz+b^2zx+c^2xy \right ) \le (a^2+b^2+c^2)(xy+zx+yz) \\\\ 2\left ( a^2yz+b^2zx+c^2xy \right ) \le a^2x(y+z)+b^2y(x+z)+c^2z(x+y)=-a^2x^2-b^2y^2-c^2z^2\leq 0\]Thus \[a^2yz+b^2zx+c^2xy \leq 0\]

 

[Albert1]

Thankyou Euge for the edit!My suggested solution:

Spoiler

WLOG, I can choose $x,y$ and $z$ such, that: $yz \ge zx \ge xy$. On the other hand, I can also choose the sides in a triangle, such that: $a^2 \le b^2 \le c^2$.

Applying Chebyshev´s sum inequality:\[3\left ( a^2yz+b^2zx+c^2xy \right ) \le (a^2+b^2+c^2)(xy+zx+yz) \\\\ 2\left ( a^2yz+b^2zx+c^2xy \right ) \le a^2x(y+z)+b^2y(x+z)+c^2z(x+y)=-a^2x^2-b^2y^2-c^2z^2\leq 0\]Thus \[a^2yz+b^2zx+c^2xy \leq 0\]

Spoiler

in fact $3\left ( a^2yz+b^2zx+c^2xy \right ) \le (a^2+b^2+c^2)(xy+zx+yz)\leq 0$
for $ xy+yz+zx\leq 0$
hence
$\left ( a^2yz+b^2zx+c^2xy \right ) \leq 0$

 

[Albert1]

Let $a,\,b$ and $c$ be the sides of a triangle and $x,\,y$ and $z$ are real numbers such that $x+ y+ z = 0$. Prove that $a^2yz +b^2xz+c^2xy\le 0$.

my solution:

Spoiler

$\because (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=0\\
\therefore yz+zx+xy\leq 0----(1)\\
(1)\times a^2:a^2yz+a^2zx+a^2xy\leq 0---(2)\\
(1)\times b^2:b^2yz+b^2zx+b^2xy\leq 0---(3)\\
(1)\times c^2:c^2yz+c^2zx+c^2xy\leq 0---(4)\\
(2)+(3)+(4): (a^2+b^2+c^2)(yz+zx+xy)\leq 0---(5)\\$
now we may set $yz\geq zx\geq xy$
and $a^2\leq b^2\leq c^2 (or \,\,a \leq b\leq c)$
applying Chebyshev's sum inequality we get:
$3(a^2yz+b^2zx+c^2xy)\leq (a^2+b^2+c^2)(yz+zx+xy)\leq 0$
that is :
$a^2yz+b^2zx+c^2xy\leq 0$

 

[kaliprasad]

my solution:

Spoiler

$\because (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=0\\
\therefore yz+zx+xy\leq 0----(1)\\
(1)\times a^2:a^2yz+a^2zx+a^2xy\leq 0---(2)\\
(1)\times b^2:b^2yz+b^2zx+b^2xy\leq 0---(3)\\
(1)\times c^2:c^2yz+c^2zx+c^2xy\leq 0---(4)\\
(2)+(3)+(4): (a^2+b^2+c^2)(yz+zx+xy)\leq 0---(5)\\$
now we may set $yz\geq zx\geq xy$
and $a^2\leq b^2\leq c^2 (or \,\,a \leq b\leq c)$
applying Chebyshev's sum inequality we get:
$3(a^2yz+b^2zx+c^2xy)\leq (a^2+b^2+c^2)(yz+zx+xy)\leq 0$
that is :
$a^2yz+b^2zx+c^2xy\leq 0$

Spoiler

applying law of symmetry you can set
either $yz\geq zx\geq xy$ or $a^2\leq b^2\leq c^2 (or \,\,a \leq b\leq c)$ but not both as a gets tied to yz , b to xz , c to xy and any random combinations

 

[Albert1]

Spoiler

applying law of symmetry you can set
either $yz\geq zx\geq xy$ or $a^2\leq b^2\leq c^2 (or \,\,a \leq b\leq c)$ but not both as a gets tied to yz , b to xz , c to xy and any random combinations

Spoiler

sure ! with different assumption,we may obtain distinct result

 

[anemone]

Sorry folks! Apparently my solution doesn't work as I have somehow misinterpreted the problem, which was silly mistake from my part...

Sorry too for my late reply as I have been pretty busy for both family and personal issue!

 

[lfdahl]

Spoiler

applying law of symmetry you can set
either $yz\geq zx\geq xy$ or $a^2\leq b^2\leq c^2 (or \,\,a \leq b\leq c)$ but not both as a gets tied to yz , b to xz , c to xy and any random combinations

You´re right, kaliprasad, therefore my suggested solution in #8 is not valid!