$\frac{a}{c}+ \frac{c}{b}= \frac{\frac{a^2}{c}}{a}+ \frac{c}{b}\ge \frac{(\frac{a}{\sqrt{c}}+\sqrt{c})^2}{a+b}\ge \frac{4a}{a+b}$

parmenides51 wrote: Let $a, b$ and $c$ denote positive real numbers. Prove that $\frac{a}{c}+\frac{c}{b}\ge \frac{4a}{a + b}$ . When does equality hold? (Walther Janous) By AM-GM. $$\frac{a}{c}+\frac{c}{b}=\frac{a^2}{ca}+\frac{c}{b}\ge \frac{2a}{\sqrt{ab}}\ge \frac{4a}{a + b}$$h

$$\frac{a}{c}+\frac{c}{b}\ge \frac{4a}{a + b}$$$$ \iff a^2b+ab^2+ac^2+bc^2 \geq 4abc$$$$ \iff a(b-c)^2+b(a-c)^2 \geq 0$$which is clearly true. Equality holds for $a=b=c$.

It is $$\frac{a}{c}+\frac{c}{b}\geq \frac{a^2}{ac}+\frac{c^2}{bc}\geq \frac{(a+c)^2}{c(a+b)}\geq \frac{4a}{a+b}$$by the TITUS lemma.

The inequality transforms to the form $a^2b+b^2a+c^2a+c^2b$>=$4abc$ which is just am gm

Expanding and simplifying, we have $$\frac{ab+c^2}{bc} \ge \frac{4a}{a+b}$$ $$a^2b+ab^2+ac^2+bc^2 \ge 4abc.$$ Now we use the AM-GM inequality; $$\frac{a^2b+ab^2+ac^2+bc^2}{4} \ge \sqrt[4]{a^2b*ab^2*ac^2*bc^2}$$ $$a^2b+ab^2+ac^2+bc^2 \ge 4 \sqrt[4]{a^4b^4c^4}$$ $$a^2b+ab^2+ac^2+bc^2 \ge 4abc.$$ Therefore, the inequality is true. Equality holds when $a=b=c.$

Solved with AI216, v4913, CT17, CyclicISLclesTrapezoid, kvedula2004 at 4. We have $$\frac{a}{c} + \frac{c}{b} \geq 2\sqrt{\frac{a}{b}} = \frac{4a}{2\sqrt{ab}} \geq \frac{4a}{a+b}$$ which implies that equality holds iff $a=b=c$.

Let $a, b$ and $c$ denote positive real numbers. Prove that $$\frac{a}{c}+\frac{kc}{b}\ge \frac{4\sqrt k a}{a + b}$$Where $k>0.$

My solution

Let $a, b$ and $c$ denote positive real numbers. Prove that $$\frac{a}{c}+\frac{24c}{b}\ge \frac{25a}{a+b+3c}$$$$\frac{a}{c}+\frac{c}{b}\ge \frac{100a}{3(a+b+24c)}$$$$\frac{a}{c}+\frac{8c}{b}\ge \frac{(2+5\sqrt 5)a}{a+b+c}$$$$\frac{a}{c}+\frac{6c}{b}\ge \frac{(5\sqrt[3]{2} +4\sqrt[3]{4}-1)a}{a+b+c}$$

a/c+c/b≥2√a/b( By AM - GM inequality) (a+b)²≥4ab a(a+b)²≥4a²b 4a/b(a+b)²≥16a² 2√a/b≥4a/(a+b) So, we can conclude that a/c+c/b≥4a/(a+b)

By, AM-GM, it follows, $\frac{a}{c}+\frac{c}{b} \ge 2\sqrt{\frac{a}{c}\cdot \frac{c}{b}}=2\sqrt{\frac{a}{b}}.$ Now, note that, $2\sqrt{\frac{a}{b}}=\frac{4a}{2\sqrt{ab}}$, because, if we square, both sides, we have $4\cdot \frac{a}{b}=\frac{16a^2}{4ab}$, hence, $16ab \cdot \frac{a}{b}=16a^2$, now multiplying, both sides by $b$, finishes. Now, note that $\frac{4a}{2\sqrt{ab}}\ge \frac{4a}{a+b}$, which follows, by AM-GM. And, we are done, equality, holds if $a=b=c.$ note that the last step is proceeded because $a+b\ge 2\sqrt{ab}$, by AM-GM, but since this is in the denominator, we have $\frac{4a}{2\sqrt{ab}}$, is bigger