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let $a,b>0$, and such $$\dfrac{2}{a}+\dfrac{1}{b}=1$$

Find this minimum $$a+b+\sqrt{a^2+b^2}$$

My try: since $$2b+a=ab$$ so $$a+b+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+2ab}+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+4b+2a}+\sqrt{a^2+b^2}$$

then I can't

maybe this problem can use AM-GM or Cauchy-Schwarz inequality solve it.Thank you very much

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Are you forbidding calculus? –  user27126 Dec 5 '13 at 6:22
    
Can we use $a=2\sin^2\theta,b=\cos^2\theta$ –  lab bhattacharjee Dec 5 '13 at 6:33
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5 Answers 5

up vote 4 down vote accepted

since $$\dfrac{1}{a}+\dfrac{2}{b}=1$$ then a straight line $$\dfrac{x}{a}+\dfrac{y}{b}=1$$ cross $P(1,2)$ enter image description here then $$a+b+\sqrt{a^2+b^2}=|OA|+|OB|+|AB|$$

In the follow enter image description here

we have $$|OC|+|OD|+|CD|\ge|OE|+|OF|+|EF|=|OA|+|OB|+AB|=10$$

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I agree that the problem is: find a triangle with vertices $0,A,B$ such that $A \in y^+$, $B \in x^+$, $P(1,2) \in \overline{AB}$ and the circumference is minimal. But I do not understand the second picture: why is the angle at $0$ not $90^\circ$, how do you get the center $Q$ of the circle and why is $|OE|+|OF|+|EF|=|OA|+|OB|+AB|$ –  miracle173 Dec 5 '13 at 8:19
    
I think the angle at $0$ is $90^\circ$ and $Q$ is the intersection of that angle bisection in $0$ and a normal to $AB$ in $P$. But $|OE|+|OF|+|EF|=|OA|+|OB|+AB|$ is still unclear to me. –  miracle173 Dec 5 '13 at 8:27
    
@miracle the angle is 90. The perimeter of either triangle is twice the distance from O to the tangency point of the circle. –  Calvin Lin Dec 5 '13 at 15:05
    
+1 very nice. You should explain how to get the circle and why it works. –  Calvin Lin Dec 5 '13 at 15:08
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Now that someone has worked out the place where the minimum is achieved ($a:b = 4:3$), one can purposely use Cauchy-Schwarz/AM-GM.

$$(a^2+b^2)(16+9) \ge (4a+3b)^2 \Rightarrow \sqrt{a^2+b^2} \ge \frac{4a+3b}{5}$$ Then $$a+b+\sqrt{a^2+b^2} \ge \frac{9a+8b}{5}$$ Again Cauchy-Schwarz, $$(9a+8b)(\frac{4}{a} + \frac{2}{b}) \ge (6+4)^2 = 100 \Rightarrow 9a+8b \ge 50$$ Therefore $$a+b+\sqrt{a^2+b^2} \ge \frac{9a+8b}{5} \ge 10$$

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let $$a=\dfrac{2(x+y)}{x},b=\dfrac{x+y}{y},x,y>0$$ then $$C=a+b+\sqrt{a^2+b^2}=\dfrac{(x+y)(x+2y+\sqrt{4y^2+x^2})}{xy}$$ so $$C-10=\dfrac{x^2-7xy+2y^2+(x+y)\sqrt{x^2+4y^2}}{xy}$$ and note $$(x+y)^2(x^2+4y^2)-(x^2-7xy+2y^2)^2=4xy(2x-3y)^2\ge 0$$ so $$a+b+\sqrt{a^2+b^2}\ge 10$$

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As you stated, $2b + a = ab$. We can get this even nicer: $a = \frac{2b}{b - 1}$.

Substitute into the square root expression: $$ \begin{align*} a + b + \sqrt{a^2 + b^2} &= \frac{2b}{b - 1} + b + \sqrt{\left( \frac{2b}{b - 1} \right)^2 + b^2} \\ &= \frac{b^2 + b}{b - 1} + \sqrt{ \frac{b^4 - 2b^3 + 5b^2}{(b - 1)^2} } \\ &= \frac{b^2 + b}{b - 1} + \frac{b}{b - 1} \sqrt{ b^2 - 2b + 5 } \\ &= \frac{b}{b - 1} (b + 1 + \sqrt{ b^2 - 2b + 5 }) \\ \end{align*} $$

From here I don't see any other way to find the minimum but calculus. Take the derivative and set equal to zero. But it's really messy. On the other hand, Wolfram Alpha gives a minimum of $\frac{5}{2}$ for it.

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Not that messy since equalling the numerator to 0 is enough to find the minimum: $b+1+\sqrt{b^2-2b+5} + b + \frac{2b^2-2b}{2\sqrt{b^2-2b+5}}) = 0$. –  András Hummer Dec 5 '13 at 11:49
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Lagrange multipliers seems like a good approach. Solve the system of equations $$\displaystyle \frac{\partial }{\partial a}\left(a+b+\sqrt{a^2+b^2}-\lambda \left(\frac{2}{a}+\frac{1}{b}-1\right)\right)=0\textrm,$$, $$\frac{\partial }{\partial b}\left(a+b+\sqrt{a^2+b^2}-\lambda \left(\frac{2}{a}+\frac{1}{b}-1\right)\right)=0\textrm,$$,

$$\frac{\partial }{\partial \lambda }\left(a+b+\sqrt{a^2+b^2}-\lambda \left(\frac{2}{a}+\frac{1}{b}-1\right)\right)=0\textrm.$$

Mathematica gives the solution as $\left(\lambda = -10, a= \frac{10}{3},b= \frac{5}{2}\right)$, which gives the minimum as $$a+b+\sqrt{a^2+b^2}= \frac{10}{3}+\frac{5}{2}+\sqrt{\left(\frac{10}{3}\right)^2+\left(\frac{5}{2}\right)^2} = 10\textrm.$$

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