Spanija 2012

Let $a$, $b$, $c>0$ and $a\cdot b\cdot c=1$.

Suppose $a+b+c>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.

Note that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}=bc+ac+ab$ since $abc=1$.

So the inequality becomes:

$$a+b+c>ab+bc+ca$$

Let us consider the possible cases for $a$, $b$, $c$.

Since $abc=1$, at least one of $a$, $b$, $c$ is $\ge 1$ and at least one is $\le 1$.

Suppose two of them are greater than $1$, say $a>1$, $b>1$, $c<1$ (since $abc=1$).

Let $a=x>1$, $b=y>1$, $c=z<1$, $xyz=1$.

Then $z=\frac{1}{xy}$.

Now, $$a+b+c=x+y+\frac{1}{xy}$$ $$ab+bc+ca=xy+y\cdot \frac{1}{xy}+x\cdot \frac{1}{xy}=xy+\frac{y}{xy}+\frac{x}{xy}=xy+\frac{1}{x}+\frac{1}{y}$$ So the inequality becomes: $$x+y+\frac{1}{xy}>xy+\frac{1}{x}+\frac{1}{y}$$ But $x>1$, $y>1$, so $xy>1$.

Let us check if this is possible.

Let $x=y=t>1$, $z=\frac{1}{t^2}<1$.

Then: $$a+b+c=t+t+\frac{1}{t^2}=2t+\frac{1}{t^2}$$ $$ab+bc+ca=t^2+t\cdot \frac{1}{t^2}+t\cdot \frac{1}{t^2}=t^2+\frac{t}{t^2}+\frac{t}{t^2}=t^2+\frac{2}{t}$$ So the inequality is: $$2t+\frac{1}{t^2}>t^2+\frac{2}{t}$$ Let us check for $t=2$: $$2\cdot 2+\frac{1}{4}=4.25$$ $$2^2+\frac{2}{2}=4+1=5$$ So $4.25<5$, so the inequality does not hold.

Try $t=1.5$: $$2\cdot 1.5+\frac{1}{(1.5{)}^2}=3+\frac{1}{2.25}\approx 3.444$$ $$(1.5{)}^2+\frac{2}{1.5}=2.25+1.333=3.583$$ So $3.444<3.583$, again the inequality does not hold.

Try $t=3$: $$2\cdot 3+\frac{1}{9}=6.111$$ $$9+\frac{2}{3}=9.666$$ So $6.111<9.666$.

So, if two numbers are greater than $1$, the inequality does not hold.

Now, suppose all three are less than $1$. Then $abc<1$, contradiction.

Suppose all three are greater than $1$. Then $abc>1$, contradiction.

Suppose exactly one is greater than $1$, say $a>1$, $b<1$, $c<1$.

Let $a=x>1$, $b=y<1$, $c=z<1$, $xyz=1$.

Then $yz=\frac{1}{x}$.

Let $y=s<1$, $z=\frac{1}{xs}<1$.

Now, $$a+b+c=x+s+\frac{1}{xs}$$ $$ab+bc+ca=xs+s\cdot \frac{1}{xs}+x\cdot \frac{1}{xs}=xs+\frac{1}{x}+\frac{1}{s}$$ So the inequality is: $$x+s+\frac{1}{xs}>xs+\frac{1}{x}+\frac{1}{s}$$ Let us try $x=2$, $s=0.5$, $z=\frac{1}{2\cdot 0.5}=1$.

But $z=1$, so two numbers are $\le 1$, one is $>1$.

Try $x=2$, $s=0.4$, $z=\frac{1}{2\cdot 0.4}=1.25$ (but $z>1$). So $z$ must be $<1$.

Try $x=2$, $s=0.6$, $z=\frac{1}{2\cdot 0.6}=0.833$.

So $a=2$, $b=0.6$, $c=0.833$.

Compute: $$a+b+c=2+0.6+0.833=3.433$$ $$ab+bc+ca=2\cdot 0.6+0.6\cdot 0.833+0.833\cdot 2=1.2+0.5+1.666=3.366$$ So $3.433>3.366$, the inequality holds.

Try $x=3$, $s=0.3$, $z=\frac{1}{3\cdot 0.3}=1.111$ (not $<1$).

Try $x=1.5$, $s=0.7$, $z=\frac{1}{1.5\cdot 0.7}\approx 0.952$.

$a=1.5$, $b=0.7$, $c=0.952$.

$a+b+c=1.5+0.7+0.952=3.152$

$ab+bc+ca=1.5\cdot 0.7+0.7\cdot 0.952+0.952\cdot 1.5=1.05+0.666+1.428=3.144$

So $3.152>3.144$, the inequality holds.

Therefore, the inequality holds only when exactly one of $a$, $b$, $c$ is greater than $1$.

Thus, if $a+b+c>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, then exactly one of $a$, $b$, $c$ is greater than $1$.