I'll call expression of this form "innocent" $$\frac{M}{x^2+y^2+z^2} + \frac{N}{xy+yz+zx}$$ if we apply some inequality (like AM-GM, Bunyakovsky,...) and we still preserve the equality at $a=b=c=\text{some value}$ What is the condition of $M$ and $N$ to make the expression up there become innocent? There is a question in my test that has the expression of that form: $$\text{Given $x+y+z=1$, prove that: }\frac{1}{x^2+y^2+z^2}+\frac{3}{xy+yz+zx}\ge12$$ I start applying AM-GM and the result is NOT true, I start to become panicking in the last 15 minutes, and it cost me 2/20 pts. Although I know the equality happens at is $x=y=z=\frac{1}{3}$, but I can't prove it. (Source: Exam for excellent students in grade 8 in Vietnam) $$\text{1: No DIFFERENTIATING (I'm a grade 8 student)}$$ $$\text{2: Of course $x,y,z$ must be POSITIVE to apply anything}$$

  • $\begingroup$ $(-1,-1,3)$ gives $ \frac{1}{11} + \frac{3}{-5}= \frac{-28}{55}$ $\endgroup$
    – Will Jagy
    Jun 24, 2022 at 1:55
  • 1
    $\begingroup$ @WillJagy I think he missed the condition $x,y,z>0$. $\endgroup$
    – ling
    Jun 24, 2022 at 1:59
  • $\begingroup$ of course $x,y,z$ must be positive $\endgroup$
    – user1070711
    Jun 25, 2022 at 6:22

1 Answer 1


First note that $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$, if we denote $t=xy+yz+zx$, we have $$\frac{1}{x^2+y^2+z^2}+\frac{3}{xy+yz+zx}=\frac{1}{1-2t}+\frac{3}{t}$$ since $x+y+z=1$. By $xy+yz+zx\leq\frac{1}{3}(x+y+z)^2=\frac{1}{3}$, we know it suffices to minimize $$f(t):=\frac{1}{1-2t}+\frac{3}{t},\quad0<t\leq\frac{1}{3}.$$ Differentiating directly yields $$f'(t)=\frac{2}{(1-2t)^2}-\frac{1}{t^2}=\frac{-10t^2+12t-3}{(1-2t)^2t^2}<0,\quad 0<t\leq\frac{1}{3}.$$ Hence we know $f(t)\geq f(\frac{1}{3})=\frac{1}{1-\frac{2}{3}}+9=12$. Then we are done!

(PS: It is easy to see that " = " holds iff $x=y=z=\frac{1}{3}$.)

  • $\begingroup$ Sorry, but no differentiating here, I'm just a grade 8 student $\endgroup$
    – user1070711
    Jun 25, 2022 at 6:24

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