let $x,y,z$ be real numbers, and such $$x+y+z+xyz=4$$ show that $$x^2+y^2+z^2+3\ge 2(xy+yz+xz)$$
My try: let $$x+y+z=p,xy+yz+xz=q,xyz=r$$ then $$p+r=4$$ then $$\Longleftrightarrow p^2+3\ge 4q$$
But I can't.Thank you
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What if $x=y=z=1$? – Michael Hoppe Jan 27 '14 at 12:16
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Please ignore ... – Michael Hoppe Jan 27 '14 at 12:25
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This is a well known problem, I did it 10 years ago, but I am not able to reproduce the proof. – Ma Ming Jan 27 '14 at 12:25
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1Problems of this kind all become routine using Lagrange multipliers. – J.R. Jan 27 '14 at 12:28
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First, one may suppose $x, y, z >0$. Use $\sin^2 A+\sin^2 B +\sin^ 2 C+2\sin A\sin B\sin C=1$ for $A+B+C=\frac{\pi}{2}$ to turn it to a trigonometric inequality. – Ma Ming Jan 27 '14 at 12:43
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Is $R$ a ring or is $R = \mathbb{R}$ the set of all real numbers? – Keba Jan 27 '14 at 13:18
3 Answers
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Note that it is sufficient to consider $x, y, z \ge 0$. Then the condition yields $4 = x+y+z+xyz \ge 3\sqrt[3]{xyz}+xyz \implies xyz \le 1$.
Also note that among $x-1, y-1, z-1$, at least two have the same sign. WLOG let $(y-1)(z-1) \ge 0$. Then we have: $$(x-1)^2+(y-z)^2+2x(y-1)(z-1) \ge 0$$ $$\implies x^2+y^2+z^2+2xyz + 1 \ge 2(xy+yz+zx)$$
which along with $xyz \le 1$ gives you the result.
Macavity
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I think you should consider $f(x,y,z): {\bf R}^3 \to {\bf R}$ defined as
$$(x,y,z) \mapsto x^2 + y^2 + z^2 -2xy -2yz -2xz + 3.$$
Then $f \in {\mathcal C}^\infty({\bf R}^3)$ and you could investigate where $f(x,y,z) = 0$. That could get you a little further.
David
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Following the suggestion by TooOldForMath given in the comments above we may define $$ f(x,y,z)=x^2+y^2+z^2+3-2(xy+yz+xz) $$ which is the multivariate function that we want to show has minimal value greater than or equal to zero when subject to the constraint $$ g(x,y,z)=x+y+z+xyz=4 $$ so we define $$ \Lambda(x,y,z,\lambda)=f(x,y,z)+\lambda(g(x,y,z)-4) $$ and calculate the corresponding system of equations $$ \begin{align} \frac{\partial\Lambda}{\partial x}&=2x-2y-2z+\lambda(1+yz)&=0\\ \frac{\partial\Lambda}{\partial y}&=2y-2x-2z+\lambda(1+xz)&=0\\ \frac{\partial\Lambda}{\partial z}&=2z-2x-2y+\lambda(1+xy)&=0\\ \frac{\partial\Lambda}{\partial \lambda}&=x+y+z+xyz-4&=0 \end{align} $$ This system is symmetrical in $x,y$ and $z$ and can be solved by Gauss elimintation to get the solutions $$ x=y=z=\lambda=1\\ \mbox{and}\\ x=4,y=z=-0.5,\lambda=-8 $$ In the last solution one may interchange $x,y$ and $z$ to get three essentially identical solutions. Moreover there is a complex valued solution too, but that is of no relevance to the current question. Plugging in $(x,y,z)=(1,1,1)$ and $(x,y,z)=(4,-0.5,-0.5)$ respectively we obtain the minimum and maximum of the function, namely $$ \begin{align} f(1,1,1)&=0\\ f(4,-0.5,-0.5)&=27 \end{align} $$ Proving that $f$ indeed is greater than or equal to zero everywhere on the curve defined by $g(x,y,z)=4$. In fact $g(x,y,z)=4\implies f(x,y,z)\in[0,27]$. The claim follows!
