sqing wrote:

wdym by that

Let $a$, $b$, $c$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} =6.$ Show that$$\frac{a^2+b}{a+b} + \frac{b^2+c}{b+c} + \frac{c^2+a}{c+a} \geq \frac{9}{4}.$$Solution: $$\sum \frac{a^2+b}{a+b} \geq\sum \frac{a+b-\frac{1}{4}}{a+b} =3-\frac{1}{4}\sum \frac{1}{a+b} \geq 3-\frac{1}{16}\sum \left(\frac{1}{a}+\frac{1}{b}\right)=\frac{9}{4}.$$ Let $a$, $b$, $c$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{9}{2}.$ Show that$$\frac{a^2+b}{a+b} + \frac{b^2+c}{b+c} + \frac{c^2+a}{c+a} \geq \frac{5}{2}.$$

Let $a$, $b$, $c$ be positive real numbers such that $\frac{1}{a^{n-1}}+\frac{1}{b^{n-1}}+\frac{1}{c^{n-1}} =\frac{3n^{n-1}}{(n-1)^{n-1}} .$ Show that$$\frac{a^n+b^{n-1}}{a+b^{n-1}} + \frac{b^n+c^{n-1}}{b^{n-1}+c^{n-1}} + \frac{c^n+a^{n-1}}{c^{n-1}+a^{n-1}} \geq \frac{3(2n-1)}{2n}.$$Where $2\leq n\in N.$ My solution: $$\sum \frac{a^n+b^{n-1}}{a^{n-1}+b^{n-1}} \geq\sum \frac{a^{n-1}+b^{n-1}-\frac{(n-1)^{n-1}}{n^n}}{a+b} =3-\frac{(n-1)^{n-1}}{n^n}\sum \frac{1}{a^{n-1}+b^{n-1}}$$$$ \geq 3-\frac{(n-1)^{n-1}}{4n^n}\sum \left(\frac{1}{a^{n-1}}+\frac{1}{b^{n-1}}\right)=\frac{3(2n-1)}{2n}.$$

Let $a$, $b$, $c$ be positive real numbers such that $a+b+c=1.$ Show that$$\frac{a^2+b}{a+b} + \frac{b^2+c}{b+c} + \frac{c^2+a}{c+a} \geq 2.$$

wrong solution

iman007 wrote: sqing wrote: Let $a$, $b$, $c$ be positive real numbers such that $a+b+c=1.$ Show that$$\frac{a^2+b}{a+b} + \frac{b^2+c}{b+c} + \frac{c^2+a}{c+a} \geq 2.$$ note that: \[\sum \frac{a}{a+b} \ge \frac32\]so we have to prove that: \[\sum \frac{a^2}{a+b} \stackrel{?}{\ge} \frac12\]which is straight forward titu. No.

sqing wrote: No. It is a nice inequality. Sorry for the lack of attention here is the corrected proof First note that by C-S we have: \[\sum \frac{a^2+b}{a+b} \ge \frac{(\sum \sqrt{a^2+b})^2}{2}\] It suffices to show that: \[\sum \sqrt{a^2+b} \ge 2 \quad (1)\] on the other hand we have that $\sum \sqrt{a^2+b}=\sum \sqrt{a^2+b(\sum a)}$ also: \[2\sum \sqrt{a^2+b^2+ab+bc}\sqrt{b^2+c^2+ca+cb} \ge 2(\sum a^2 +3\sum ab)\] which implies: \[\sum \sqrt{a^2+b(\sum a)} \ge 2(\sum a) \quad (2)\] now considering the fact that $a+b+c=1$ by just putting it in $(2)$ we instantly get the desired result.