Let $y=x+1, z=x+2$. Then we must have $\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2} > \frac{1}{45}$. Obviously $\frac{3}{x} > \frac{1}{45} \implies x < 135$. To maximize $x+y+z = 3x+3$, we maximize $x$. If we let $x=134$, then $\frac{1}{134}+\frac{1}{135}+\frac{1}{136} > \frac{1}{45}$, which is true because this is equivalent with \[\frac{\frac{1}{134}+\frac{1}{136}}{2} > \frac{1}{135}\] which is true because the inequality \[\frac{1}{x-1}+\frac{1}{x+1} > \frac{2}{x} \implies \frac{2x}{x^2-1} > \frac{2}{x} \implies x^2>x^2-1\] is true, so we're done. Therefore the maximum is $\boxed{405}$.

Puerto Rico Math Olympiad 2010 Find all real number $r$ such that the inequality $r(bc+ca+ab)+(3-r)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9$ holds for all positive real numbers $a,b,c.$

sqing wrote: Puerto Rico Math Olympiad 2010 Find all real number $r$ such that the inequality $r(bc+ca+ab)+(3-r)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9$ holds for all positive real numbers $a,b,c.$ is the answer $r=1$ and eq. holds if $a=b=c=1$

sqing wrote: Puerto Rico Math Olympiad 2010 Find all real number $r$ such that the inequality $r(bc+ca+ab)+(3-r)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9$ holds for all positive real numbers $a,b,c.$ This is straightforward Holder's (?) which gives

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sqing wrote: Puerto Rico Math Olympiad 2010 Find all real number $r$ such that the inequality $r(bc+ca+ab)+(3-r)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9$ holds for all positive real numbers $a,b,c.$ 2007 Bulgarian Autumn Math Competition

sqing wrote: sqing wrote: Puerto Rico Math Olympiad 2010 Find all real number $r$ such that the inequality $r(bc+ca+ab)+(3-r)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9$ holds for all positive real numbers $a,b,c.$ 2007 Bulgarian Autumn Math Competition Would anyone kindly provide a solution to this inequality when $r = 1$? My progress (which becomes faulted at the end): $$ab +bc +ac + 2\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \geq 9$$We have $$\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)(ab+bc+ac) \geq (a+b+c)^2$$so $$\iff (ab+bc+ac)+2 \cdot \frac{(a+b+c)^2}{ab+bc+ac} \geq 9$$AM-GM: $$(ab+bc+ac) + 2 \cdot \frac{(a+b+c)^2}{ab+bc+ac} \geq 3\sqrt{2}(a+b+c)$$And the above is incorrect.

Let $x=y-1$ and $z=y+1$ and $n=45$. Note that $x+y+z=3y$. We claim that $9n=\boxed{405}$ is the maximum. It is achieved when $y=3n$, since: $$\frac1x+\frac1y+\frac1z-\frac1{45}=\frac1{3n-1}+\frac1{3n}+\frac1{3n+1}-\frac1n=\frac2{3n(9n^2-1)}>0.$$Now suppose $y\ge3n+1$. Then: $$\frac1x+\frac1y+\frac1z\le\frac1{3n}+\frac1{3n+1}+\frac1{3n+2}\le\frac1{3n}+\frac1{3n}+\frac1{3n}=\frac1{45},$$which contradicts $\frac1x+\frac1y+\frac1z>\frac1{45}$.

Write $x,z=y\pm1$. The optimal construction is $y=135$, which yields $3\cdot135=\boxed{405}$. Proof of validity. We require $\tfrac{1}{134}+\tfrac{1}{135}+\tfrac{1}{136}>\tfrac{1}{45}$, or equivalently, $\tfrac{1}{134}+\tfrac{1}{136}>\tfrac{1}{135}+\tfrac{1}{135}$. Since $134+136=135+135$, it suffices to show $134\cdot136<135\cdot135$, which is immediate by difference-of-squares. Proof of optimality. Clearly $f(y)=\tfrac{1}{y-1}+\tfrac{1}{y}+\tfrac{1}{y+1}$ is monotonically decreasing, so proving $\tfrac{1}{135}+\tfrac{1}{136}+\tfrac{137}>\tfrac{1}{45}$ is sufficient. But $\tfrac{1}{45}=\tfrac{1}{135}+\tfrac{1}{135}+\tfrac{135}{135}$, and now the result is apparent. $\blacksquare$