Suppose $2b-\frac{1}{c}$, $2a-\frac{1}{b}$, $2b-\frac{1}{c}$ and $2c-\frac{1}{a} >2$ Multiply and expanding: $8abc-\frac{1}{abc}-4(a+b+c)+\frac{2}{a}+\frac{2}{b}+\frac{2}{c}>8$ Summing up and expanding: $2a+2b+2c-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}>6$ Denote $A=2a+2b+2c-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}$. Then $\frac{15}{4}>A>6$(Contradiction)

Let $2a-\frac{1}{b}>2,2b-\frac{1}{c}>2,2c-\frac{1}{a}>2$ Easy to get that then $a>0,b>0,c>0$ $2a-\frac{1}{b}>2 \to 2ab>2b+1$ and so $2ac>2a+1, 2bc>2c+1$ but then $32=8(abc)^2> (2a+1)(2b+1)(2c+1) \geq ( 2\sqrt[3]{abc}+1)^3=8abc+6\sqrt[3]{abc}+12\sqrt[3]{(abc)^2}+1>16+6+12+1=35$

My solution during the test: Note $abc=2$ implies either all are positive or exactly one is positive. If exactly one is positive, say $a,$ then $2c-\frac{1}{a}<0<2,$ so we must have all positive. Now, $2a-\frac{1}{b}>2$ implies $b>\frac{1}{2a-2},$ which implies $c<\frac{4a-4}{a}.$ Now, $2c-\frac{1}{a}>2$ implies $a>1.5,$ so $abc>1.5^3>2.$ Took me embarrassingly long (25 minutes)

I think we need to show that it is possible for two numbers to be larger than 2 after proving it is not possible for three numbers to be larger than 2.

PikaNiko wrote: I think we need to show that it is possible for two numbers to be larger than 2 after proving it is not possible for three numbers to be larger than 2. Well, we need to prove that At most two of them are greater than 2. We only show the example if they ask us to prove that exactly two of them are greater than 2

PikaNiko wrote: Suppose $a$, $b$ and $c$ are nonzero real numberss satisfying $abc=2$. Prove that among the three numbers $2a-\frac{1}{b}$, $2b-\frac{1}{c}$ and $2c-\frac{1}{a}$, at most two of them are greater than $2$. https://artofproblemsolving.com/community/c1068820h2938799p26300984 https://artofproblemsolving.com/community/c6h511348p2871225

Suppose $a$, $b$ , $c$ and $d$ are nonzero real numbers satisfying $abcd=2$. Prove that among the three numbers $2a-\frac{1}{bc}$, $2b-\frac{1}{cd}$ , $2c-\frac{1}{da}$and $2d-\frac{1}{ab}$, at most three of them are greater than $2$.

Schur-Schwartz wrote: Suppose $2b-\frac{1}{c}$, $2a-\frac{1}{b}$, $2b-\frac{1}{c}$ and $2c-\frac{1}{a} >2$ Multiply and expanding: $8abc-\frac{1}{abc}-4(a+b+c)+\frac{2}{a}+\frac{2}{b}+\frac{2}{c}>8$ Summing up and expanding: $2a+2b+2c-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}>6$ Denote $A=2a+2b+2c-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}$. Then $\frac{15}{4}>A>6$(Contradiction) That is almost the same as what I wrote Sadly, it took me about 45 mins to figure it out. It is not very hard tho

Suppose $a$, $b$ , $c$ and $d$ are nonzero real numbers satisfying $abcd=2$. Among the four numbers $2a-\frac{1}{b}$, $2b-\frac{1}{c}$ , $2c-\frac{1}{d}$ and $2d-\frac{1}{a}$ at most there are several greater than $2$?

A straightforward approach. Let $WLOG$, $a=\frac{2x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$ Assuming they’re all large than two and substituting We find $4x>2y+z, 2y>2z+x, 4z>4x+y$, its easy to prove this is impossible.

PikaNiko wrote: Suppose $a$, $b$ and $c$ are nonzero real numberss satisfying $abc=2$. Prove that among the three numbers $2a-\frac{1}{b}$, $2b-\frac{1}{c}$ and $2c-\frac{1}{a}$, at most two of them are greater than $2$. let's assume contrary $2a>2+\frac{1}{b}$ $2b>2+\frac{1}{c}$ $2c>2+\frac{1}{a}$ then it must be $8abc=16>(2+\frac{1}{c})(2+\frac{1}{a})(2+\frac{1}{b})$ but by $AM$$\ge$$GM$ $(2+\frac{1}{c})(2+\frac{1}{a})(2+\frac{1}{b})$$\ge$$8×\sqrt{8×\frac{1}{abc}}$$=$$16$ contradiction