Let $a,b,c \in (1, \infty)$ such that $ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}=2$. Prove that: $$ \sqrt {a-1} + \sqrt {b-1} + \sqrt {c-1} \leq \sqrt {a+b+c}. $$ This is supposed to be solved using the Cauchy inequality; that is, the scalar product inequality.
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We apply the Cauchy Schwarz Inequality to the vectors $x = \left({\sqrt {\dfrac{a-1}{a}} , \sqrt {\dfrac{b-1}{b}} , \sqrt {\dfrac{c-1}{c}}}\right) $ and $y = \left({\dfrac{1}{\sqrt{bc}},\dfrac{1}{\sqrt{ac}}, \dfrac{1}{\sqrt{ab}} }\right)$ in $\Bbb R^3$.
Then, $x \cdot y \le \lVert x\rVert \lVert y\rVert$ yields,
$$ \dfrac{\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1}}{\sqrt{abc}} \le \sqrt{\dfrac{a-1}{a} + \dfrac{b-1}{b} + \dfrac{c-1}{c}} \times \sqrt{\dfrac{1}{bc} + \dfrac{1}{ac} + \dfrac{1}{ab}}$$
Now manipulating the $\sqrt{abc}$ across the sign we get
$$ \sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} \le \sqrt{\dfrac{a-1}{a} + \dfrac{b-1}{b} + \dfrac{c-1}{c}} \times \sqrt{a+ b+c} $$
So, $$ \dfrac{\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1}}{\sqrt{a+ b+c}} \le \sqrt{\left({1 - \dfrac 1 a}\right) + \left({1 - \dfrac 1 b}\right) + \left({1 - \dfrac 1 c}\right) }$$
$$ \dfrac{\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1}}{\sqrt{a+ b+c}} \le \sqrt{3 - \left({\dfrac 1 a + \dfrac 1 b + \dfrac 1 c}\right)} = \sqrt{3-2} = 1$$
$\mathscr{Q.E.D}$
Ishfaaq
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Let $a=x+1$, $b=y+1$ and $c=z+1$.
Hence, $x$, $y$ and $z$ are positives and we need to prove that $$\sqrt{x}+\sqrt{y}+\sqrt{z}\leq\sqrt{x+y+z+3}$$ or $\sqrt{xy}+\sqrt{xz}+\sqrt{yz}\leq\frac{3}{2}$.
But in another hand, the condition geves $$\sum_{cyc}\frac{1}{x+1}=2$$ or $$\sum_{cyc}\left(\frac{1}{x+1}-1\right)=-1$$ or $$\sum_{cyc}\frac{x}{x+1}=1.$$ Thus, by C-S $$1=\sum_{cyc}\frac{x}{x+1}\geq\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{x+y+z+3},$$ which gives $\sqrt{xy}+\sqrt{xz}+\sqrt{yz}\leq\frac{3}{2}$.
Done!
Michael Rozenberg
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