Given $ 3x^{2}+3xz+z^{2} = 1 \quad\quad (1)$, $ 3y^{2}+3yz+z^{2} = 4 \quad\quad (2)$ and $ x^{2}-xy+y^{2} = t \quad\quad (3)$. Calculate $4(1)+4(2)-2(3)$: $4(3x^{2}+3xz+z^{2})+4(3y^{2}+3yz+z^{2})-2(x^{2}-xy+y^{2})=4+16-2t$, then $9x^{2}+12xz+4z^{2}+9y^{2}+12yz+4z^{2}+x^{2}+2xy+y^{2}=20-2t$ or $(3x+2z)^{2}+(3y+2z)^{2}+(x+y)^{2}=20-2t$, only if $20-2t \geq 0$.

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Assmit wrote: Solution is complete? i think so

No, it is not complete I think, problem has missed condition, that $x,y,z$ are positive numbers I found papers with solution here

Solution idea

Answer