http://www.kalva.demon.co.uk/short/soln/sh9326.html

We can also solve this by Lagrange Method(a little similar to the solution given in KALVA but i think Lagrange Method is better) that i've posted about in my Blog: http://www.mathlinks.ro/weblog_entry.php?t=163432 let: $ f = abc+bcd+cda+dab-\frac{176}{27}abcd$ $ g = a+b+c+d-1 = 0$ We have to prove that always we have: $ f\leq\frac{1}{27}$ Let: $ L = f-\mu g = abc+bcd+cda+dab-\frac{176}{27}abcd-\mu a-\mu b-\mu c-\mu d+\mu$ $ \frac{\partial L}{\partial a}= bc+cd+db-\frac{176}{27}bcd-\mu = 0$ $ \frac{\partial L}{\partial b}= ac+cd+da-\frac{176}{27}acd-\mu = 0$ $ \frac{\partial L}{\partial c}= ab+bd+da-\frac{176}{27}bad-\mu = 0$ $ \frac{\partial L}{\partial d}= bc+ca+ab-\frac{176}{27}bca-\mu = 0$ $ \Longrightarrow\mu = bc+cd+db-\frac{176}{27}bcd = ac+cd+da-\frac{176}{27}acd = ab+bd+da-\frac{176}{27}bad = bc+ca+ab-\frac{176}{27}bca$ Now notice that: $ bc+cd+db-\frac{176}{27}bcd = ac+cd+da-\frac{176}{27}acd\Longrightarrow(b-a)\left(c+d-\frac{176}{27}cd\right) = 0$ by the same way we'll have the following equations too: $ (b-c)\left(a+d-\frac{176}{27}ad\right) = 0$ $ (b-d)\left(a+c-\frac{176}{27}ca\right) = 0$ $ (a-c)\left(b+d-\frac{176}{27}bd\right) = 0$ $ (a-d)\left(c+b-\frac{176}{27}cb\right) = 0$ $ (c-d)\left(a+b-\frac{176}{27}ab\right) = 0$ By solving these six equations we must have $ a = b = c = d(\text{Why?})$ and so by the condition $ a+b+c+d = 1$ we'll have $ a = b = c = d =\frac{1}{4}$ and: $ f(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4}) =\frac{1}{4^{3}}+\frac{1}{4^{3}}+\frac{1}{4^{3}}+\frac{1}{4^{3}}-\frac{176}{27}\cdot\frac{1}{4^{4}}=\frac{1}{27}$ And we're done!

Beside Lagrange Method, we can have some more elementary solution, as (1), By SMV theorem, see here (hungkhtn) http://www.mathlinks.ro/Forum/viewtopic.php?t=113248 (2), By AC (or generally known as UMV for Vietnamese) theorem, see Mathematics Reflection (VASC) (3), By EV theorem, see CRUX or Gazeta mathematica (VASC) (4), By Convex function, which is the most elementary way,

Letting $ f(a;b;c;d) = abc + bcd + cda + dab - \frac {176abcd}{27}$ $ = ab(c + d - kcd) + cd(a + b)$$ (k = \frac {176}{27})$ Hence we hope prove that $ f(a;b;c;d) \leq f(t;t;c;d)$ where $ t = \frac {a + b}{2}$.Since $ 0 \leq ab \leq t^2$ we need to prove $ c + d - kcd \geq 0$ Otherwise,if $ c + d - kcd \leq 0$,then we have $ f(a;b;c;d) = ab(c + d - kcd) + cd(a + b) \leq cd(a + b) \leq \frac {1}{27}$ Thus,we may assume that: $ f(a;b;c;d) \leq f(\frac {a + b}{2};\frac {a + b}{2};c;d)$ Put $ s = \frac {c + d}{2}$ we have: $ f(a;b;c;d) \leq f(t;t;c;d) \leq f(t;t;s;s) = f(t;s;t;s) \leq f(\frac {t + s}{2};\frac {t + s}{2};t;s) \leq f( \frac {t + 2}{2}; \frac {t + 2}{2}; \frac {t + 2}{2}; \frac {t + 2}{2}) = \frac {1}{27}$ Done! http://www.mathlinks.ro/Forum/viewtopic.php?t=168355

And the ineq with n=5? PS: I think that the LHS is not enough! It can be symmetric with the variables!

Attachments:

hmm! Noone solves it!

ricardokaka wrote: hmm! Noone solves it! Sorry ricardokaka but it is one of problems in Mathematics and Youth Magazine, for the moment we are not dicussing anything about above inequality. In my opinion, mixing variable method kills it easily

orl wrote: Let $a,b,c,d$ be four non-negative numbers satisfying \[ a+b+c+d=1. \] Prove the inequality \[ a \cdot b \cdot c + b \cdot c \cdot d + c \cdot d \cdot a + d \cdot a \cdot b \leq \frac{1}{27} + \frac{176}{27} \cdot a \cdot b \cdot c \cdot d. \] We can prove it by simple MV-method: Let $a=\min\{a,b,c,d\}$ and $f(a,b,c,d)=1+176abcd-27\sum_{cyc}abc$. Hence, $a\leq\frac{1}{4}$ and $f(a,b,c,d)-f\left(a,\frac{b+c+d}{3},\frac{b+c+d}{3},\frac{b+c+d}{3}\right)=$ $=176a\left(bcd-\left(\frac{b+c+d}{3}\right)^3\right)-27\left(a\left(bc+bd+cd-3\left(\frac{b+c+d}{3}\right)^2\right)+bcd-\left(\frac{b+c+d}{3}\right)^3\right)=$ $=\left(\left(\frac{b+c+d}{3}\right)^3-bcd\right)(27-176a)+9a(b^2+c^2+d^2-bc-bd-cd)=$ $=\left(\left(\frac{b+c+d}{3}\right)^3-bcd\right)(27-176a)+\frac{9a}{b+c+d}\cdot(b^3+c^3+d^3-3bcd)\geq$ $=\left(\left(\frac{b+c+d}{3}\right)^3-bcd\right)(27-176a)+\frac{9a}{b+c+d}\cdot\frac{27}{4}\left(\left(\frac{b+c+d}{3}\right)^3-bcd\right)=$ $=\left(\left(\frac{b+c+d}{3}\right)^3-bcd\right)\left(27-176a+\frac{243a}{4(1-a)}\right)=$ $=\left(\left(\frac{b+c+d}{3}\right)^3-bcd\right)\cdot\frac{704a^2-569a+108}{4(1-a)}\geq0$ for $a\leq\frac{1}{4}$. Id est, it remains to prove that $f(a,b,b,b)\geq0$, which after homogenization is equivalent to: $(a+3b)^4+176ab^3\geq27(a+3b)(3ab^2+b^3)$, which is $a(a+14b)(a-b)^2\geq0$. Done!

nice proof arqady,mv is effective.but the computing is terrible,

I think it's IMO shortlist. I have a nice proof. Let $a+b=x, ab=m, c+d=y, cd=n$. then $x+y=1$ we will prove that $nx+my \leq 1/27 + (176/27) mn$ But above inequality is linear function for $m,n$, and $m$ is arbitrary numbers such that $0 \leq m \leq x^2/4$, $n$ is arbitrary numbers such that $0 \leq n \leq y^2/4$. So, It's enough to prove the inequality where $m=0\; or\; x^2/4$, $n=0 \;or\; y^2/4$ It's easy!

By the same way (it's really the same way!) we can prove the following generalization. Let $\sum_{i=1}^na_i=1$, where $a_i>0$, $n\geq3$, $k_n=n^2(n-1)^{n-1}-n^n$ and $m_n=(n-1)^{n-1}$. Prove that: \[\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\leq\frac{1+k_na_1a_2...a_n}{m_na_1a_2...a_n}\] For $n=2$ this inequality is an equality.