Given $0\leq a\leq b\leq c$ and $b > 0$, prove that: $2\sqrt{\frac{a}{c}}\leq \frac{b}{c} + \frac{a}{b}\leq 1 + \frac{a}{c}$ This is an exercise that got me stuck. It's from the book "Functions of Several Real Variables 1.1 - exercise 10" by Martin Moskowitz and Fotios Paliogiannis.
Exercise:
Given $0\leq a\leq b\leq c$ and $b > 0$, prove that:
$2\sqrt{\frac{a}{c}}\leq \frac{b}{c} + \frac{a}{b}\leq 1 + \frac{a}{c}$
What I did so far:
$2\sqrt{\frac{a}{c}}\leq \frac{b}{c} + \frac{a}{b}\leq 1 + \frac{a}{c}$
$4\frac{a}{c}\leq \frac{b^2}{c^2} +2\frac{a}{c} + \frac{a^2}{b^2}\leq 1 + 2\frac{a}{c}+\frac{a^2}{c^2}$  [power of 2]
$2\frac{a}{c}\leq \frac{b^2}{c^2} + \frac{a^2}{b^2}\leq 1 +\frac{a^2}{c^2}$ [minus $2\frac{a}{c}$]
$2acb^2 \leq b^4+a^2c^2 \leq c^2b^2+a^2b^2$ [multiply by $c^2b^2$]
$2acb^2 -2a^2c^2\leq b^4\leq c^2b^2+a^2b^2 - a^2c^2$
And now I have to prove the inequalities for $b^4$. However, I can not advance (actually I have no way of knowing if this is the right path since I have very little experience writing proofs).
 A: It is easier to prove the two inequalities separately.

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*$2\sqrt{\frac{a}{c}}\leq \frac{b}{c} + \frac{a}{b}\quad$ This is just AM-GM $\,\sqrt{uv} \le \frac{u+v}{2}\,$ with $\,u=\frac{b}{c}\,$, $\,v=\frac{a}{b}\,$.


*$\frac{b}{c} + \frac{a}{b}\leq 1 + \frac{a}{c}\quad$ This is $\,1 + \frac{a}{c} -\frac{b}{c} - \frac{a}{b} \ge 0 \iff \left(\frac{b}{c} - \frac{a}{c}\right)\left(\frac{c}{b} - 1\right) \ge 0\,$.

[ EDIT ] $\;$ This works the same with the inequalities on the last-but-one line in OP's post.

$2acb^2 \leq b^4+a^2c^2 \leq c^2b^2+a^2b^2$


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*$2acb^2 \leq b^4+a^2c^2\iff (b^2 - ac)^2 \ge 0\,$ which always holds.


*$b^4+a^2c^2 \leq c^2b^2+a^2b^2$ $\iff c^2b^2+a^2b^2-b^4-a^2c^2 \ge 0$ $\iff (b^2 - a^2) (c^2 - b^2) \ge 0\,$ which holds because $\,c \ge b \ge a \ge 0\,$.
A: Let $a=x^2, b=y^2, c=z^2.$ For fixed $b,c$ let $f(x)=(b/c)+(a/b)-2\sqrt {a/c}\,=(y^2/z^2)+(x^2/y^2)-2(x/z)$ and let $g(a)=1+(a/c)-(b/c)-(a/b).$
Show that $f'(x)>0\iff x>y^2/z.$ So $\min f(x)=f(y^2/z)=0.$
Show that $g'(a)\le 0.$ So $\min \{g(a):a\le b\}=g(b)=0.$
The substitution $a=x^2, b=y^2, c=z^2$ is not really necessary. It just makes the calculations easier for the left inequality.
