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Question Number 7883 by 314159 last updated on 23/Sep/16
Let a,b,c be the lengths of the sides of a triangle.  Show that abc≥(a+b−c)(b+c−a)(c+a−b).
$${Let}\:{a},{b},{c}\:{be}\:{the}\:{lengths}\:{of}\:{the}\:{sides}\:{of}\:{a}\:{triangle}. \\ $$$${Show}\:{that}\:{abc}\geqslant\left({a}+{b}−{c}\right)\left({b}+{c}−{a}\right)\left({c}+{a}−{b}\right). \\ $$
Commented by sou1618 last updated on 23/Sep/16
  x=a+b−c>0  y=c+a−b>0    a,b,c :length of triangle  z=b+c−a>0    x+y≥2(√(xy))>0  y+z≥2(√(yz))>0  z+x≥2(√(zx))>0    (x+y)(y+z)(z+x)≥8xyz  (2a)×(2c)×(2b)≥8(a+b−c)(c+a−b)(b+c−a)  abc≥(a+b−c)(b+c−a)(c+a−b) .
$$ \\ $$$${x}={a}+{b}−{c}>\mathrm{0} \\ $$$${y}={c}+{a}−{b}>\mathrm{0}\:\:\:\:{a},{b},{c}\::{length}\:{of}\:{triangle} \\ $$$${z}={b}+{c}−{a}>\mathrm{0} \\ $$$$ \\ $$$${x}+{y}\geqslant\mathrm{2}\sqrt{{xy}}>\mathrm{0} \\ $$$${y}+{z}\geqslant\mathrm{2}\sqrt{{yz}}>\mathrm{0} \\ $$$${z}+{x}\geqslant\mathrm{2}\sqrt{{zx}}>\mathrm{0} \\ $$$$ \\ $$$$\left({x}+{y}\right)\left({y}+{z}\right)\left({z}+{x}\right)\geqslant\mathrm{8}{xyz} \\ $$$$\left(\mathrm{2}{a}\right)×\left(\mathrm{2}{c}\right)×\left(\mathrm{2}{b}\right)\geqslant\mathrm{8}\left({a}+{b}−{c}\right)\left({c}+{a}−{b}\right)\left({b}+{c}−{a}\right) \\ $$$${abc}\geqslant\left({a}+{b}−{c}\right)\left({b}+{c}−{a}\right)\left({c}+{a}−{b}\right)\:. \\ $$
Commented by sou1618 last updated on 23/Sep/16
when...    abc=(a+b−c)(b+c−a)(c+a−b)    ⇔x=y=z    ⇔ { ((a+b−c=b+c−a)),((b+c−a=c+a−b)),((c+a−b=a+b−c)) :}    ⇔a=b=c
$${when}… \\ $$$$\:\:{abc}=\left({a}+{b}−{c}\right)\left({b}+{c}−{a}\right)\left({c}+{a}−{b}\right) \\ $$$$\:\:\Leftrightarrow{x}={y}={z} \\ $$$$\:\:\Leftrightarrow\begin{cases}{{a}+{b}−{c}={b}+{c}−{a}}\\{{b}+{c}−{a}={c}+{a}−{b}}\\{{c}+{a}−{b}={a}+{b}−{c}}\end{cases} \\ $$$$\:\:\Leftrightarrow{a}={b}={c} \\ $$
Commented by Rasheed Soomro last updated on 23/Sep/16
Nice!
$$\mathcal{N}{ice}! \\ $$
Answered by prakash jain last updated on 02/Oct/16
see comments
$$\mathrm{see}\:\mathrm{comments} \\ $$

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