If $a,b,c,d>0$ and $a+b=3$
and $\displaystyle \frac{a}{c}+\frac{b}{d}=1$. Then
$\min{(c+d)}$ is
I have used Cauchy Inequality
$\displaystyle (\frac{a}{c}+\frac{b}{d})\cdot (c+d)\geq (\sqrt{a}+\sqrt{b})^2$
$c+d\geq (\sqrt{a}+\sqrt{b})^2=3+2\sqrt{ab}$
How I find minimum value of $c+d$
Please have a look on it
Best Answer
There is no answer. choosing $a=\varepsilon, b=3-\varepsilon, c=\varepsilon +\sqrt{3\varepsilon -\varepsilon ^2}, d=\sqrt{\dfrac{3-\varepsilon}{\varepsilon}}\cdot \left( \varepsilon +\sqrt{3\varepsilon -\varepsilon ^2} \right)$
$$\frac{a}{c}+\frac{b}{d}=\frac{\varepsilon}{\varepsilon +\sqrt{3\varepsilon -\varepsilon ^2}}+\frac{\sqrt{3\varepsilon -\varepsilon ^2}}{\varepsilon +\sqrt{3\varepsilon -\varepsilon ^2}}=1 $$
And: $$c+d=\left( \varepsilon +\sqrt{3\varepsilon -\varepsilon ^2} \right) \left( 1+\sqrt{\frac{3-\varepsilon}{\varepsilon}} \right) =\left( \sqrt{\varepsilon}+\sqrt{3-\varepsilon} \right) ^2=3+2\sqrt{3\varepsilon -\varepsilon ^2}\rightarrow 3 $$
Meaning that the minimum doesn't exist (And that the best infimum is $3$).
How to quickly do the construction:
Suppose $\dfrac{a}{c}=x\in(0,1)$, then $\dfrac{b}{d}=1-x\Rightarrow c=\dfrac{a}{x}, d=\dfrac{b}{1-x}\Rightarrow c+d=\dfrac{a}{x}+\dfrac{3-a}{1-x} \geq \dfrac{(\sqrt{a}+\sqrt{3-a})^2}{1}$ Using the Cauchy-Schwarz inequality, and also note that the '$=$' can always be obtained when $$\dfrac{a}{x^2}=\dfrac{3-a}{\left( 1-x \right) ^2}\Rightarrow \left( \dfrac{1}{x}-1 \right) ^2=\dfrac{3}{a}-1(>0)$$
Therefore we only need to analyze $(\sqrt{a}+\sqrt{3-a})^2=3+2\sqrt{a(3-a)}$, but apparently by choosing $a$ sufficiently close to $0$ or $3$ leads to the fact that there is only an infimum $c+d>3$.