Prooove it!
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a,b,c are positive real numbers.\[abc=1\]Prove that,\[\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\geq \frac{3}{2}\]
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- nafistiham
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Re: Prooove it!
WLOG let us assume $c\leq b\leq a$ let $x= \frac {1}{a},y=\frac {1}{b},z=\frac {1}{c}$ now,
now, we can find the left hand side,
\[\frac{x^{3}}{\frac{1}{y}+\frac{1}{z}}+\frac{y^{3}}{\frac{1}{x}+\frac{1}{z}}+\frac{z^{3}}{\frac{1}{x}+\frac{1}{y}}\]
\[=\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y}\]
since $x\leq y \leq z$ we can deduce that, $x+y\leq x+z \leq y+z$ and also that, $\frac{x}{y+z}\leq \frac{y}{x+z}\leq \frac{z}{x+y}$
\[\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \geq \frac{xy}{y+z}+ \frac{yz}{x+z}+ \frac{zx}{x+y}\]
\[\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \geq \frac{xz}{y+z}+ \frac{yx}{x+z}+ \frac{zy}{x+y}\]
if we add both sides we'll get
\[2 \left ( \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right) \geq x+y+z \geq 3\sqrt[3]{xyz} = 3\]
therefore,
\[\frac{1}{a^{3}(b+c)}+\frac{1} {b^{3}(c+a)}+\frac{1}{c^{3}(b+c)} \geq \frac{3}{2} \]
now, we can find the left hand side,
\[\frac{x^{3}}{\frac{1}{y}+\frac{1}{z}}+\frac{y^{3}}{\frac{1}{x}+\frac{1}{z}}+\frac{z^{3}}{\frac{1}{x}+\frac{1}{y}}\]
\[=\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y}\]
since $x\leq y \leq z$ we can deduce that, $x+y\leq x+z \leq y+z$ and also that, $\frac{x}{y+z}\leq \frac{y}{x+z}\leq \frac{z}{x+y}$
\[\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \geq \frac{xy}{y+z}+ \frac{yz}{x+z}+ \frac{zx}{x+y}\]
\[\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \geq \frac{xz}{y+z}+ \frac{yx}{x+z}+ \frac{zy}{x+y}\]
if we add both sides we'll get
\[2 \left ( \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right) \geq x+y+z \geq 3\sqrt[3]{xyz} = 3\]
therefore,
\[\frac{1}{a^{3}(b+c)}+\frac{1} {b^{3}(c+a)}+\frac{1}{c^{3}(b+c)} \geq \frac{3}{2} \]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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- Nadim Ul Abrar
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Re: Prooove it!
$ab+bc+ca\geq 3\sqrt{abc}=3$ [AM-GM]
$\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}=\frac{(bc)^2}{a(b+c)}+\frac{(ca)^2}{b^3(c+a)}+\frac{(ab)^2}{c(a+b)} \geq \frac{(ab+bc+ca)^2}{2(ab+bc+ca)}=\frac{ab+bc+ca}{2} \geq \frac{3}{2} $[ C.S Engel ]
$\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}=\frac{(bc)^2}{a(b+c)}+\frac{(ca)^2}{b^3(c+a)}+\frac{(ab)^2}{c(a+b)} \geq \frac{(ab+bc+ca)^2}{2(ab+bc+ca)}=\frac{ab+bc+ca}{2} \geq \frac{3}{2} $[ C.S Engel ]
$\frac{1}{0}$
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Re: Prooove it!
I didn't know it was posted before...by the way..thanks to help.bro. Can i know how much time did u spend to prove it?nafistiham wrote:WLOG let us assume $c\leq b\leq a$ let $x= \frac {1}{a},y=\frac {1}{b},z=\frac {1}{c}$ now,
now, we can find the left hand side,
\[\frac{x^{3}}{\frac{1}{y}+\frac{1}{z}}+\frac{y^{3}}{\frac{1}{x}+\frac{1}{z}}+\frac{z^{3}}{\frac{1}{x}+\frac{1}{y}}\]
\[=\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y}\]
since $x\leq y \leq z$ we can deduce that, $x+y\leq x+z \leq y+z$ and also that, $\frac{x}{y+z}\leq \frac{y}{x+z}\leq \frac{z}{x+y}$
\[\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \geq \frac{xy}{y+z}+ \frac{yz}{x+z}+ \frac{zx}{x+y}\]
\[\frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \geq \frac{xz}{y+z}+ \frac{yx}{x+z}+ \frac{zy}{x+y}\]
if we add both sides we'll get
\[2 \left ( \frac{x^{2}}{y+z}+\frac{y^{2}}{x+z}+\frac{z^{2}}{x+y} \right) \geq x+y+z \geq 3\sqrt[3]{xyz} = 3\]
therefore,
\[\frac{1}{a^{3}(b+c)}+\frac{1} {b^{3}(c+a)}+\frac{1}{c^{3}(b+c)} \geq \frac{3}{2} \]
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.
- nafistiham
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Re: Prooove it!
well, actually i did it during the inequality camp going on the group some days ago.as, the task was to learn most part of book "Inequalities", i used the examples to do the exercises.
and this is one of the solved examples of that book.
and this is one of the solved examples of that book.
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
- Phlembac Adib Hasan
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