[不等式] 晨早流流证个RJ简单不等式$ab+bc+ca=3$

已知 $a,b,c>0,ab+bc+ca=3$，求证\[
\frac1{1+a^2(b+c)}+\frac1{1+b^2(c+a)}+\frac1{1+c^2(a+b)}\leqslant\frac1{abc}
\]
问题来自 http://bbs.pep.com.cn/thread-1899855-1-1.html

注意到\[
\frac1{1+a^2(b+c)}=\frac1{1+a(ab+ca)}=\frac1{1+a(3-bc)}=\frac1{1+3a-abc}
\]又显然\[
abc=\sqrt{ab\cdot bc\cdot ca}\leqslant\sqrt{\left(\frac{ab+bc+ca}{3}\right)^3}=1
\]所以\[
\frac1{1+a^2(b+c)}=\frac1{1+3a-abc}\leqslant\frac1{3a}
\]另外两项同理，故\[
\frac1{1+a^2(b+c)}+\frac1{1+b^2(c+a)}+\frac1{1+c^2(a+b)}\leqslant\frac1{3a}+\frac1{3b}+\frac1{3c}
=\frac{ab+bc+ca}{3abc}=\frac1{abc}
\]

那这个呢？
For positive real numbers$a,b$  and $c$ we have$a+b+c=3 $ . Prove that:
\begin{align}
\frac{a}{1+(b+c)^{2}}+\frac{b}{1+(a+c)^{2}}+\frac{c}{1+(a+b)^{2}}\le\frac{3(a^{2}+b^{2}+c^{2})}{a^{2}+b^{2}+c^{2}+12abc}
\end{align}

完全改头换面了……

Cauchy-Schwarz kills it!

那这个呢？
For positive real numbers$a,b$  and $c$ we have$a+b+c=3 $ . Prove that:
\begin{align}
\frac{a}{1+(b+c)^{2}}+\frac{b}{1+(a+c)^{2}}+\frac{c}{1+(a+b)^{2}}\le\frac{3(a^{2}+b^{2}+c^{2})}{a^{ ...
pxchg1200 发表于 2011-10-8 16:02

(Vo Quoc Ba Can):
Write the inequality as
\[{\left[ a-\frac{a}{1+(b+c)^2}\right] +\left[b-\frac{b}{1+(c+a)^2}\right]+\left[ c-\frac{c}{1+(a+b)^2}\right] \ge 3-\frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc},}\] \[\frac{a(b+c)^2}{1+(b+c)^2}+\frac{b(c+a)^2}{1+(c+a)^2}+\frac{c(a+b)^2}{1+(a+b)^2} \ge \frac{36abc}{a^2+b^2+c^2+12abc}.\]
Using the Cauchy-Schwarz inequality, we get
\[\left[ \sum \frac{a(b+c)^2}{1+(b+c)^2}\right] \left\{ \sum \frac{a\left[1+(b+c)^2\right]}{(b+c)^2}\right\} \ge \left(\sum a\right)^2 =9.\]

Therefore, it suffices to prove that
\[\frac{9}{\frac{a\left[1+(b+c)^2\right]}{(b+c)^2}+\frac{b\left[1+(c+a)^2\right]}{(c+a)^2}+\frac{c\left[1+(a+b)^2\right]}{(a+b)^2}} \ge \frac{36abc}{a^2+b^2+c^2+12abc},\] or \[\frac{a^2+b^2+c^2+12abc}{abc} \ge 4\left\{\frac{a\left[1+(b+c)^2\right]}{(b+c)^2}+\frac{b\left[1+(c+a)^2\right]}{(c+a)^2}+\frac{c\left[1+(a+b)^2\right]}{(a+b)^2}\right\}.\]
Since
\[\begin{aligned}
\frac{a\left[1+(b+c)^2\right]}{(b+c)^2}+\frac{b\left[1+(c+a)^2\right]}{(c+a)^2}+\frac{c\left[1+(a+b)^2\right]}{(a+b)^2}&=\left[ \frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}\right]+(a+b+c) \\ &= \frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}+3,
\end{aligned} \]
the above inequality can be written as
\[\frac{a^2+b^2+c^2}{abc}  \ge 4\left[  \frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}\right] ,\]
which is true because
\[4\left[  \frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}\right] \le 4\left(\frac{a}{4bc}+\frac{b}{4ca}+\frac{c}{4ab}\right) =\frac{a^2+b^2+c^2}{abc}.\]
The proof is completed. $\blacksquare$

5# pxchg1200


矮油……关键还是那个柯西了……