hopefully my last question for proofing (1 Viewer)

AbstractBlade · Mar 28, 2021

This is hopefully my last question for proofing

if x,y,z $\in$ real numbers and $x \neq y \neq z$
i) prove $x^2 + y^2 + z^2 > yz + zx + xy$
ii) if $x + y + z = 1, show yz + zx + xy < 1/3$

CM_Tutor · Mar 28, 2021

Both of these are standard questions following from AMGM for n=3

[edit] - actually, easier still... see below

AbstractBlade · Mar 28, 2021

CM_Tutor said:
Both of these are standard questions following from AMGM for n=3

thank you

CM_Tutor · Mar 28, 2021

Or also:

$(a-b)^2 \geqslant 0 \qquad \implies a^2 + b^2 \geqslant 2ab \qquad \qquad \text{. . . . . . . . . (*)}$

Putting $a = x$ and $b = y$ into (*) yields $x^2 + y^2 \geqslant 2xy$

Putting $a = y$ and $b = z$ into (*) yields $y^2 + z^2 \geqslant 2yz$

Putting $a = z$ and $b = x$ into (*) yields $z^2 + x^2 \geqslant 2zx$

Summing these three inequalities proves the first result

CM_Tutor · Mar 28, 2021

Then, noting that $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$

and using this with $x+y+z=1$ in the above and the previous result should get the answer needed

AbstractBlade · Mar 28, 2021

CM_Tutor said:
Then, noting that $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$

and using this with $x+y+z=1$ in the above and the previous result should get the answer needed

thank you, you have helped alot

CM_Tutor · Mar 28, 2021

CM_Tutor said:
Or also:

$(a-b)^2 \geqslant 0 \qquad \implies a^2 + b^2 \geqslant 2ab \qquad \qquad \text{. . . . . . . . . (*)}$

Putting $a = x$ and $b = y$ into (*) yields $x^2 + y^2 \geqslant 2xy$

Putting $a = y$ and $b = z$ into (*) yields $y^2 + z^2 \geqslant 2yz$

Putting $a = z$ and $b = x$ into (*) yields $z^2 + x^2 \geqslant 2zx$

Summing these three inequalities proves the first result

CM_Tutor said:
Then, noting that $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$

and using this with $x+y+z=1$ in the above and the previous result should get the answer needed

Note that the AM-GM-HM inequalities yield equality only if all terms are the same.

Thus, if I know (say) that $x \neq y$ then I can start with $(x-y)^2 > 0$ as the LHS can only be zero if $x-y=0 \implies x=y$ .

So, in both questions here, applying the restriction that $x \neq y \neq z$ makes the inequalities strict.

hopefully my last question for proofing (1 Viewer)

AbstractBlade

New Member

CM_Tutor

Moderator

AbstractBlade

New Member

CM_Tutor

Moderator

CM_Tutor

Moderator

AbstractBlade

New Member

CM_Tutor

Moderator

Users Who Are Viewing This Thread (Users: 0, Guests: 1)