Dadgarnia wrote: If $0<a\leq b\leq c$ prove that $$\frac{(c-a)^2}{6c}\leq \frac{a+b+c}{3}-\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ \[Q=\frac{a+b+c}{3}-\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}-\frac{(c-a)^2}{6c}\] $Q=\frac{1}{6}\,{\frac { \left( a-2\,b+c \right) ^{2}bc+2\, \left( a-b \right) ^{ 2} \left( b-a \right) c+6\, \left( a-b \right) ^{2} \left( c-b \right) c+ \left( a-c \right) ^{2} \left( c-b \right) a}{c \left( ab+ ca+bc \right) }}\geq 0 $

Dadgarnia wrote: If $0<a\leq b\leq c$ prove that $$\frac{(c-a)^2}{6c}\leq \frac{a+b+c}{3}-\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ Simple Inequality. If $0<a\leq b\leq c$ . Prove that $$\frac{a+b+c}{3}\ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+\frac{(c-a)^2}{6c}$$

Dadgarnia wrote: If $0<a\leq b\leq c$ prove that $$\frac{(c-a)^2}{6c}\leq \frac{a+b+c}{3}-\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ We have \[RHS-LHS=\frac{a(c+a-2b)^2}{6(ab+bc+ca)}+\frac{(c-a)(a-b)^2}{3(ab+bc+ca)}+\frac{b(c-a)^3}{6c(ab+bc+ca)}\ge{0}\]

((c-a)^2)÷6c《((c-a)^2)÷(3a+3c) ☆☆☆☆ ☆☆ ((c-a)^2)÷(3a+3c)《((a+b+c)÷3)-(3÷((1÷a)+(1÷b)+(1÷c)))====>>>(2ac-ab-bc)^2》0

For #5 we need to prove that that $$\frac{a+b+c}{3}\ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+\frac{(c-a)^2}{3(c+a)}$$

Right after expanding we must prove that : $$ 2ac{(c-b)^2} +{2c^2}{( b-a)^2} + (bc-ab-ac){(c-a)^2} \ge \ 0 $$we write $$ bc-ab-ac= (a-b)*(a-c)-a^2 $$so we find that the above inequality is reduced to: $$ (b-a)(c-a) {(c-a)^2} +2ac{(c-b)^2}+{2c^2} {( b-a)^2}- {a^2}{(c-a)^2} \ge \ 0 $$if we write $$ c-a = c-b+ b-a $$one can see that we receive the inequality: $$ (b-a)(c-a) {(c-a)^2} +(ac-a^2){(c-b)^2}+(c^2-a^2) {( b-a)^2}+ac{(c-b)^2}+{c^2} {( b-a)^2} - 2(c-b)(b-a){a^2} \ge \ 0 $$now if we prove $$ ac{(c-b)^2}+{c^2} {( b-a)^2} - 2(c-b)(b-a){a^2} \ge \ 0 $$we are done , which is indeed an obvious AM-GM , since we know $0<a\leq b\leq c$

here is my solution: https://www.photobox.co.uk/my/photo/full?photo_id=9174019732

here is another proof:

Attachments:

Consider the function $$ f(b)=\frac{a+b+c}{3}-\frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}-\frac{(c-a)^2}{6c}. $$We want to prove that this function always takes positive values. So we need to check for extrema points and beginning and end of the interval that $b$ is defined on; Which is $b=a$, $b=c$ and wherever $f'(b)$ is equal to $0$. We have: $$ f'(b)=\frac{1}{3}+\frac{3ac(\sum ab) - (a+c)(3abc)}{(\sum ab)^2}. $$Putting that equal to zero we'll get: $$ (\sum ab)^2 = 9ac\sum ab -9abc(a+c) = 9a^2c^2 $$$$ \Rightarrow \sum ab = 3ac \Rightarrow 2ac=ab+bc $$$$ \Rightarrow b= \frac{2}{\frac{1}{a}+\frac{1}{c}}.$$Checking $b=a$, $b=c$ and $b=\frac{2}{\frac{1}{a}+\frac{1}{c}}$ will give us that $f(b)$ is always positive as desired.

xzlbq wrote: Dadgarnia wrote: If $0<a\leq b\leq c$ prove that $$\frac{(c-a)^2}{6c}\leq \frac{a+b+c}{3}-\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ \[Q=\frac{a+b+c}{3}-\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}-\frac{(c-a)^2}{6c}\] $Q=\frac{1}{6}\,{\frac { \left( a-2\,b+c \right) ^{2}bc+2\, \left( a-b \right) ^{ 2} \left( b-a \right) c+6\, \left( a-b \right) ^{2} \left( c-b \right) c+ \left( a-c \right) ^{2} \left( c-b \right) a}{c \left( ab+ ca+bc \right) }}\geq 0 $ How should you find that

sqing wrote: For #5 we need to prove that that $$\frac{a+b+c}{3}\ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+\frac{(c-a)^2}{3(c+a)}$$ The proof, here https://www.facebook.com/photo.php?fbid=7766269253389502&set=p.7766269253389502&type=3