Question

if a,b,c >0 then a/b+c + b/c+a + c/a+b >=k.. Then the value of k is?

Answer 1

By AM - GM inequality we have,

$\small\text{ \longrightarrow\dfrac{\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)}{3}\geq\left(\dfrac{a}{b+c}\cdot\dfrac{b}{c+a}\cdot\dfrac{c}{a+b}\right)^{\frac{1}{3}} }$

$\small\text{ \longrightarrow\dfrac{\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)}{3}\geq\left(\dfrac{abc}{(a+b)(b+c)(c+a)}\right)^{\frac{1}{3}}\quad\dots(1) }$

Now the term $\small\text{ \left(\dfrac{abc}{(a+b)(b+c)(c+a)}\right)^{\frac{1}{3}} }$ has to be maximum.

Again by AM - GM inequality we have the following.

• $\small\dfrac{a+b}{2}\geq\sqrt{ab}\quad\dots(2)$

• $\small\dfrac{b+c}{2}\geq\sqrt{bc}\quad\dots(3)$

• $\small\dfrac{c+a}{2}\geq\sqrt{ca}\quad\dots(4)$

Multiplying (2), (3) and (4) we get,

$\small\longrightarrow\dfrac{a+b}{2}\cdot\dfrac{b+c}{2}\cdot\dfrac{c+a}{2}\geq\sqrt{a^2b^2c^2}$

$\small\longrightarrow\dfrac{(a+b)(b+c)(c+a)}{8}\geq abc$

$\small\longrightarrow\dfrac{abc}{(a+b)(b+c)(c+a)}\leq\dfrac{1}{8}$

$\small\text{ \longrightarrow\left(\dfrac{abc}{(a+b)(b+c)(c+a)}\right)^{\frac{1}{3}}\leq\left(\dfrac{1}{8}\right)^{\frac{1}{3}} }$

$\small\text{ \longrightarrow\left(\dfrac{abc}{(a+b)(b+c)(c+a)}\right)^{\frac{1}{3}}\leq\dfrac{1}{2} }$

For maximum,

$\small\text{ \longrightarrow\left(\dfrac{abc}{(a+b)(b+c)(c+a)}\right)^{\frac{1}{3}}=\dfrac{1}{2} }$

Then (1) becomes,

$\small\text{ \longrightarrow\dfrac{\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)}{3}\geq\dfrac{1}{2} }$

$\small\longrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}$

Therefore,

$\small\text{ \longrightarrow\underline{\underline{k=\dfrac{3}{2}}} }$