# Some easy inequality for a new day :)))

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) $$\text{1: No DIFFERENTIATING (I'm a grade 8 student)}$$ $$\text{2: Of course x,y,z must be POSITIVE to apply anything}$$

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

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 $$f(t):=\frac{1}{1-2t}+\frac{3}{t},\quad0 Differentiating directly yields $$f'(t)=\frac{2}{(1-2t)^2}-\frac{1}{t^2}=\frac{-10t^2+12t-3}{(1-2t)^2t^2}<0,\quad 0 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 $$x=y=z=\frac{1}{3}$$.)