MHB Inequality Challenge: Prove 3x2y2+x2z2+y2z2 ≤ 3

[lfdahl]

Prove, that

\[3(x^2y^2+x^2z^2+y^2z^2)-2xyz(x+y+z) \leq 3,\: \: \: \forall x,y,z \in \left [ 0;1 \right ]\]

 

[Albert1]

Prove, that

\[3(x^2y^2+x^2z^2+y^2z^2)-2xyz(x+y+z) \leq 3,\: \: \: \forall x,y,z \in \left [ 0;1 \right ]\]

my solution:

Spoiler

let :$A=3(x^2y^2+x^2z^2+y^2z^2)-2xyz(x+y+z) \leq 3(x^4+y^4+z^4)-6(\sqrt[3]{x^4y^4z^4})=B$
for each $x,y,z\in [0;1], B\geq 3\sqrt[3]{x^4y^4z^4}$
when:
$x=y=z=1$
$A=B=3$

 

[lfdahl]

my solution:

Spoiler

let :$A=3(x^2y^2+x^2z^2+y^2z^2)-2xyz(x+y+z) \leq 3(x^4+y^4+z^4)-6(\sqrt[3]{x^4y^4z^4})=B$
for each $x,y,z\in [0;1], B\geq 3\sqrt[3]{x^4y^4z^4}$
when:
$x=y=z=1$
$A=B=3$

Good job, Albert! Thankyou for the solution.Solution by other:

Spoiler

\[3(x^2y^2+x^2z^2+y^2z^2)-2xyz(x+y+z)= \\\\x^2y^2+x^2z^2+y^2z^2 + 2xy^2z-2x^2yz-2xyz^2 \\\\x^2y^2+x^2z^2+y^2z^2 + 2xyz^2-2x^2yz-2xy^2z \\\\x^2y^2+x^2z^2+y^2z^2 + 2x^2yz-2xy^2z-2xyz^2= \\\\(xy+yz-xz)^2+(xz+yz-xy)^2+(xy+xz-yz)^2\]

Now, since

\[xy+yz-xz = (x+z)y-xz \leq (x+z)-xz = (x-1)(1-z)+1\leq 1\]

and

\[xy+yz-xz \geq -xz \geq -1\]

we have

\[(xy+yz-xz)^2 \leq 1\]
By the same way$(xz+yz-xy)^2 \leq 1$ and $(xy+xz-yz)^2 \leq 1$, and we´re done.

 

[Albert1]

my solution:

Spoiler

let :$A=3(x^2y^2+x^2z^2+y^2z^2)-2xyz(x+y+z) \leq 3(x^4+y^4+z^4)-6(\sqrt[3]{x^4y^4z^4})=B$
for each $x,y,z\in [0;1], B\geq 3\sqrt[3]{x^4y^4z^4}$
when:
$x=y=z=1$
$A=B=3$

another solution:

Spoiler

$x,y\in [0;1]\\
let:
0\leq A=xy+yz+zx\leq 3\\
0\leq B=x^2y^2+y^2z^2+z^2x^2\leq 3\\
0\leq C=xyz(x+y+z)\leq 3$
we have:$0\leq A^2=B+2C\leq 9$
now :$3B-2C\leq K---(*)$
we will find $max(K)>0$ satisfying $(*)$
$\rightarrow 0\leq 3B\leq 2C+K\leq 9$
for $x,y,z\in [0;1]$ , $A,B,C$ are increasing , it is clear $K=3$,
and we get $3(x^2y^2+y^2z^2+z^2x^2)-2xyz(x+y+z)\leq 3$

 

[lfdahl]

another solution:

Spoiler

$x,y\in [0;1]\\
let:
0\leq A=xy+yz+zx\leq 3\\
0\leq B=x^2y^2+y^2z^2+z^2x^2\leq 3\\
0\leq C=xyz(x+y+z)\leq 3$
we have:$0\leq A^2=B+2C\leq 9$
now :$3B-2C\leq K---(*)$
we will find $max(K)>0$ satisfying $(*)$
$\rightarrow 0\leq 3B\leq 2C+K\leq 9$
for $x,y,z\in [0;1]$ , $A,B,C$ are increasing , it is clear $K=3$,
and we get $3(x^2y^2+y^2z^2+z^2x^2)-2xyz(x+y+z)\leq 3$

Once again: Very good job, Albert!