AoPS Academy
, where a, b, and c are positive integers, what is the smallest value of a+b+c?
and 
be positive real numbers such that 
. Prove that![\[
\frac{x^{3}}{(1 + y)(1 + z)}+\frac{y^{3}}{(1 + z)(1 + x)}+\frac{z^{3}}{(1 + x)(1 + y)}
\geq \frac{3}{4}.
\]](http://latex.artofproblemsolving.com/5/6/5/56568d27a296f1c900fabbb1042549cd2e552e29.png)
and be positive real numbers such that . Prove that ![\[\frac{x^{n}}{(\lambda + y)(\lambda+ z)}+\frac{y^{n}}{(\lambda + z)(\lambda + x)}+\frac{z^{n}}{(\lambda + x)(\lambda + y)}
\geq \frac{3}{(\lambda+1)^2}. \]](http://latex.artofproblemsolving.com/4/3/9/4395b6b3db2c363d76dfd38944073bbac78be271.png)
(
)
be three positive real numbers such that . Show that:![\[ \displaystyle \frac{x}{(1+x)(1+y)}+\frac{y}{(1+y)(1+z)}+ \frac{z}{(1+z)(1+x)} \geq \frac{3}{4}. \]](http://latex.artofproblemsolving.com/6/d/1/6d1cb9fcdc96edba81b95d11e18003de06846c7a.png)
and be positive real numbers such that . Prove that ()![\[\sum\limits_{cyc} {\frac{{{x^n}}}{{(\lambda + y)(\lambda + z)}}} \ge \frac{{{{(x + y + z)}^n}}}{{{3^{n - 2}}\left[ {\sum {(\lambda + y)(\lambda + z)} } \right]}} \quad \quad\quad\quad(1)\]](http://latex.artofproblemsolving.com/e/d/a/eda558628e94987f1ba36da67e56ccde730c9cb2.png)
But 
and using 
it suffices to show that ![\[{(\lambda + 1)^2}{(x + y + z)^n} \ge {3^{n - 1}}\left[ {3{\lambda ^2} + 2\lambda (x + y + z) + (xy + yz + zx)} \right]\quad \quad\quad\quad(2)\]](http://latex.artofproblemsolving.com/a/6/b/a6b8c3dfb9ab3bdee83a39f5e14ae534e40741ae.png)
Observe that by AM-GM using we have ![$ x + y + z \ge 3 \cdot \sqrt[3]{{xyz}} \Leftrightarrow x + y + z \ge 3$](http://latex.artofproblemsolving.com/0/a/2/0a2cec78f95570b11deee44e9228b33baddb26ec.png)
. Also we have: ![\[{\lambda ^2}{(x + y + z)^n} \ge {3^n}{\lambda ^2}\]](http://latex.artofproblemsolving.com/2/0/7/2076ef2b037ba9fdc9dc6a0c95e6e5c6a98c5043.png)
![\[2\lambda {(x + y + z)^n} = 2\lambda {(x + y + z)^{n - 1}}(x + y + z) \ge 2\lambda \cdot {3^{n - 1}}(x + y + z)\]](http://latex.artofproblemsolving.com/9/8/5/985d77cd584118da3c79e131026d0840002934eb.png)
![\[{(x + y + z)^n} = {(x + y + z)^{n - 2}}{(x + y + z)^2} \ge {3^{n - 2}} \cdot 3(xy + yz + zx) = {3^{n - 1}}(xy + yz + zx)\]](http://latex.artofproblemsolving.com/8/5/7/8571e3a8cec98d1426314473b1b98b76d52ffede.png)
Summing the above we have: ![\[{(x + y + z)^n}\left( {{\lambda ^2} + 2\lambda + 1} \right) \ge {3^{n - 1}}\left[ {{\lambda ^2} + 2\lambda (x + y + z) + (xy + yz + zx)} \right]\]](http://latex.artofproblemsolving.com/b/5/d/b5dc34e0f38879e796b853b44faec9639d41fd84.png)
We have equality only if 
.
which is equivalent to ![\[ \frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \ge \frac{(x+y+z)^3}{(3 + x + y + z)^2} \]](http://latex.artofproblemsolving.com/6/7/d/67d7a341e7d77e0fbd303690122cf422329c8fc4.png)
![\[ \frac{(x+y+z)^3}{(3 + x + y + z)^2} = \frac{(x+y+z)^2}{(x+y+z) + 6 + \frac{9}{x+y+z}} \ge \frac{(x+y+z)^2}{(x+y+z) + 9}, \]](http://latex.artofproblemsolving.com/9/0/e/90ed3e7bee13e13fe5cdccdde0d9a435bcf329bc.png)
where 
follows from 
.
such that 
.
:
.
,
and 
.
,![$\text{LHS}\geq-\frac{3}{4}+\frac{x+y+z}{2}\geq-\frac{3}{4}+\frac{3\sqrt[3]{xyz}}{2}=-\frac{3}{4}+\frac{3}{2}=\frac{3}{4}=\text{RHS}\; .$](http://latex.artofproblemsolving.com/0/2/c/02c49028d7dfbc52dea9719485fc90ef70197b68.png)
then
so 
,\[ \begin{align*} \frac{(x+y+z)^2}{(x+y+z) + 9} = \frac{(x+y+z)^2}{(x+y+z) + 9(xyz)^{1/3}} \ge \frac{(x+y+z)^2}{x+y+z + 3(x+y+z)} = \frac{(x+y+z)^2}{4(x+y+z)} = \frac{x+y+z}{4} \ge \frac{3(xyz)^{1/3}}{4} = \frac{3}{4} \end{align*} ,\] quod erat demonstrandum.
and we get
![\[\sum_{cyc} \frac{x^3}{(1+y)\cdot(1+z)}\geq \sum_{cyc} \frac{x^3}{(x+1)^2}\]](http://latex.artofproblemsolving.com/c/2/2/c22ba78c9971d3939b37b209715e9bc500cf4528.png)
.![\[\sum_{cyc} \frac{x^3}{(x+1)^2}\geq \frac{3}{4}\]](http://latex.artofproblemsolving.com/5/f/c/5fccff8a2d754b80a4350f456dd083bde89c7736.png)
.
![\[\sum_{cyc} \frac{a^3}{b(a+b)^2}\geq \sum_{cyc} \frac{a^3}{a(2a)^2}\geq \frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\]](http://latex.artofproblemsolving.com/4/d/2/4d28f4700f82c2a8f9ff193dc488ea7f5690871c.png)
.Which is true by rearrangement inequality.
![\[\text{LHS} = \sum\limits_{cyc} \dfrac{x^3}{(y + 1)(z + 1)} \ge \dfrac{\left(x + y + z\right)^3}{\left(\sum\limits_{cyc} (y + 1)\right)\left(\sum\limits_{cyc} (z + 1)\right)}\]](http://latex.artofproblemsolving.com/5/5/5/5551969d5d9d3819ab58081fd03835878f1ccdf6.png)
![\[= \dfrac{\left(x + y + z\right)^3}{\left(x + y + z + 3\right)^2}\]](http://latex.artofproblemsolving.com/2/2/8/228f264ba7d0fa460197da2d71b8e8528eb70e40.png)
)![\[= \dfrac{t^3}{\left(t + 3\right)^2} \ge \dfrac{3}{4}\]](http://latex.artofproblemsolving.com/8/d/d/8dddee7662f42fd863089789b80b514199a2aad2.png)
![\[\iff 4t^3 \ge 3\left(t + 3\right)^2\]](http://latex.artofproblemsolving.com/c/4/b/c4b462181bd6c88b6aeeae18a6199a01d372c4f5.png)
![\[\iff (t - 3)\left(4t^2 + 4t + 9\right) \ge 0\]](http://latex.artofproblemsolving.com/e/d/a/edaf6bddbb9e61e3cdd5620edc07f83578fba8a0.png)
![\[t = x + y + z \ge 3\sqrt[3]{xyz} = 3.\]](http://latex.artofproblemsolving.com/3/c/2/3c2ec10948a312f533904990fdfaf1d01e7ffdbf.png)
It would suffice if we could show 
,
So it suffices to show
Which follows from muirhead since 
,
both majorize 
.
Now we can turn this into a 
variable inequality, let 
.
Which is true if 
,
![\[\sum_{\mathrm{cyc}} x^3(1+x)\ge\frac{3}{4}\prod_{\mathrm{cyc}}(1+x).\]](http://latex.artofproblemsolving.com/1/b/d/1bd110b9aa73e2f24a92d53657c57e97af8c0568.png)
Make the substitution 
,
,
.![\[4\sum_{\mathrm{cyc}}a^9(abc+a^3)\ge 3abc\prod_{\mathrm{cyc}}(abc+a^3).\]](http://latex.artofproblemsolving.com/c/e/1/ce127ef63fd51aa238796c3b0e8b45f7871d423c.png)
Expanding, this reduces down to showing![\[4\sum_{\mathrm{cyc}}a^{10}bc + 4\sum_{\mathrm{cyc}}a^12\ge 6a^4b^4c^4 + 3\sum_{\mathrm{cyc}}a^6b^3c^3 + 3\sum_{\mathrm{cyc}}a^5b^5c^2.\]](http://latex.artofproblemsolving.com/d/2/a/d2a6593c5de449ff4cdcd53e62dbcc22de75fa83.png)
This follows from adding the following four inequalities, all of which directly follow from Murihead.
![\[(\sqrt[3]{3})\left(\sum_{cyc} (1+y)(1+z) \right)^{\frac{1}{3}} \left(\sum_{cyc} \frac{x^3}{(1+y)(1+z)} \right)^{\frac{1}{3}} \geq (x+y+z)\]](http://latex.artofproblemsolving.com/1/f/f/1ffbe7122049d86a683d2998214fa3187a37536b.png)
Cubing, it remains to prove that 
.![\[12\sum_{sym}x^2y + 4 \sum_{cyc} x^3 \geq 18(x+y+z)+ 9(xy+xz+yz)\]](http://latex.artofproblemsolving.com/4/2/a/42a3bdb08c71e4553e8b0e173b30b11355496dda.png)
Homogonizing we get from Muirhead that ![\[\sum_{sym} x^{\frac{5}{3}}y^{\frac{2}{3}}z^{\frac{-1}{3}} \geq \sum_{sym} xy\]](http://latex.artofproblemsolving.com/1/f/1/1f1d01e937d05062318023d02319819cb7e69e08.png)
![\[\sum_{sym} x^{\frac{8}{3}}y^{\frac{-1}{3}}z^{\frac{-1}{3}} \geq \sum_{sym} xy\]](http://latex.artofproblemsolving.com/f/4/8/f481e44b7ab2147310caef1c2f55c5d287b444fd.png)
![\[\sum_{sym} x^{\frac{4}{3}}y^{\frac{1}{3}}z^{\frac{-2}{3}} \geq \sum_{sym} x\]](http://latex.artofproblemsolving.com/3/9/a/39a4106ad02d160b590c817dc7db59202890c2f8.png)
![\[\sum_{sym} x^{\frac{7}{3}}y^{\frac{-2}{3}}z^{\frac{-2}{3}} \geq \sum_{sym} x\]](http://latex.artofproblemsolving.com/e/5/e/e5e82fd10337fb5d643e0ed95e8dbab1d3d3e4a2.png)
Combining all of these we can obviously see the result
means we want to prove 
We use 
Since 
,
.
,
.
Since 
by AM-GM, we just need to prove that 
.
and summing cyclically proves the desired result.
trick... should probably know to like do something with it and use simple bounds at least xd oops
be the sum. Notice![\[S\cdot \sum_{\text{cyc}}(1+y)\cdot \sum_{\text{cyc}}(1+z)\ge (x+y+z)^3\]](http://latex.artofproblemsolving.com/1/9/b/19b95dd86e555c6551dd6a0ed8ba94c1504b5211.png)
by Holder which implies![\[S\ge \frac{(x+y+z)^3}{(3+x+y+z)^2}.\]](http://latex.artofproblemsolving.com/d/e/0/de0606fa21c0eba307b8f70fd20728148cdf502c.png)
![\[\frac{s^3}{s^2+6s+9}\ge \frac{3}{4}\iff (s-3)(4s^2+9s+9)\ge 0\]](http://latex.artofproblemsolving.com/3/a/1/3a116d658f72ef110501a9aa98e15e1997d9ec02.png)
and since![\[s^3\ge 27xyz=27\implies s\ge 3\]](http://latex.artofproblemsolving.com/3/f/6/3f6d03feac6282560e779abed8760a1626522b3f.png)
we are done. 
![\[\frac{x^3}{(1+x)^2}=\frac{x^2}{\tfrac{1}{x}+2+x}\]](http://latex.artofproblemsolving.com/5/3/5/535f6a6019944d093ff7ab063d2e6714cc4d44ea.png)
and applying Titu yields something like 
.
![\[\left(\sum_\text{cyc} \frac{x^3}{(1 + y)(1 + z)}\right)(3 + x + y + z)^2 \geq (x + y + z)^3\]](http://latex.artofproblemsolving.com/0/b/0/0b0b4a97b8f3051d2ac0bb609bd44a6333bb43f1.png)
![\[ \implies \sum_\text{cyc} \frac{x^3}{(1 + y)(1 + z)} \geq \frac{(x + y + z)^3}{(3 + x + y + z)^2}.\]](http://latex.artofproblemsolving.com/3/d/3/3d37372d4cd8319489f987888366b91746cf1c6f.png)
Also, we have ![$x + y + z \geq 3\sqrt[3]{xyz} = 3$](http://latex.artofproblemsolving.com/6/c/b/6cbcb996c91689f2322a37870e0f6a537181eb1c.png)
by AM-GM. Setting 
,![\[\frac{s^3}{(3 + s)^2} \geq \frac34\]](http://latex.artofproblemsolving.com/e/6/4/e64851b7a462713aa89732fca127d71cc4a396e1.png)
for 
,
,
so proving 
is sufficient, which reduces to 
which is true after a straight AM-GM so we are done.
.
and 
where 
.
In order to prove this, consider the following inequalities which are a direct application of AM-GM inequality:
Now by applying Weighted AM-GM inequality we will get the following inequalities:
Adding these 
inequalities and dividing by 
in both the sides we shall get:
Now multiplying both the sides by 
we have,
Now finally doing 
we get,
as desired. 
So we want to show that if 
after rearranging. Note however by AM-GM that ![$s=x+y+z\ge3\sqrt[3]{xyz}=3$](http://latex.artofproblemsolving.com/3/a/c/3ac9236a72d6d0c2a63a61753d10c5d0a73cbb5c.png)
,
then it suffices to show that 
for all 
.
which can be checked, and 
,
,
,
is strictly increasing over the interval 
.
Homogenize
which is true by Muirhead.
so we're done.
.![\[\sum(a^{12}+a^9abc)=\sum(a^12+a^9)\ge \frac{3}{4}(1+a^3)(1+b^3)(1+c^3)=\frac{3}{4}abc(abc+a^3)(abc+b^3)(abc+c^3)\]](http://latex.artofproblemsolving.com/4/7/a/47ab856b1e7685dbb38b982527555cfca36c3c2c.png)
where this inequality is homogeneous.
![\[4(a^{12}+b^{12}+c^{12})+4(a^{10}bc+ab^{10}c+abc^{10})\ge 6a^4b^4c^4+3(a^6b^3c^3+a^3b^6c^3+a^3b^3c^6)+3(a^5b^5c^2+a^5b^2c^5+a^2b^5c^5)\]](http://latex.artofproblemsolving.com/5/1/c/51c7995d08159318448e3441940f0c924727a5a6.png)
and all sums can be reduced to a symmetric sum in 
,
,
,
yields 
,
.
and we find that 
.
which is true since 
by AM-GM.