This great inequality is first created by Sung Yoon Kim if I am not wrong. Nice solution by Rearrangement Inequality,

It's here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100960

Yimin Ge wrote: Let $ a,b,c,d$ be positive real numbers with $ a+b+c+d = 4$. Prove that \[ a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab\leq 4.\] Suppose that $ (x;y;z;t)$ is a permutation of $ a;b;c;d$ such that $ x\geq y\geq z\geq t$ =>$ xyz\geq xyt\geq xzt\geq yzt$ =>$ x.xyz+y.xyt+z.xyt+z.xzt+t.yzt=(xy+zt)(xz+yt)\leq\frac{1}{4}(xz+xy+yt+zt)^{2}\leq 4$ Done!

일반성을 잃지 않고 a >= b >= c >= d 라고 한다면 a*abc + b*bcd + c*cda = d*dab = abcd(a/d + b/a + c/b + d/c) < abcd(a/d + b/d + c/d + d/d) =abcd*4/d = 4abc 이제 4abc <= 4 인걸 증명해보자 AM-GM(산술기하평균에 의해) 4abc <= 4*((a + b + c)/3)^3 <= 4*((a + b + c + d)/3)^3 = 4*(4/3)^3 <= 1 증명끝

qoqqo2 wrote: 일반성을 잃지 않고 a >= b >= c >= d 라고 한다면 But our inequality is not symmetric.

By Rearrangement $ a^2bc+b^2cd+c^2da+d^2ab \le a^4+b^4+c^4+d^4$ It remains to prove that $ a^4+b^4+c^4+d^4 \le 4$ which follows from Power-Mean Inequality

hsbhatt wrote: By Rearrangement $ a^2bc + b^2cd + c^2da + d^2ab \le a^4 + b^4 + c^4 + d^4$ It remains to prove that $ a^4 + b^4 + c^4 + d^4 \le 4$ which follows from Power-Mean Inequality But it 's wrong $ (a+b+c+d)^4 \le\ 4^3(a^4 + b^4 + c^4 + d^4)$ $ \rightarrow a^4 + b^4 + c^4 + d^4 \ge 4$

oh shucks!

Here is solution using only AM-GM inequality and some algebraic manipulations.Author is Faruk Zejnulahi, teacher at Sarajevo University of Science. SOLUTION: Notice that: $ a^{2}bc + b^{2}cd + c^{2}da + d^{2}ab - (ab + cd)(ac + bd) = bd(b - d)(c - a)$ and $ a^{2}bc + b^{2}cd + c^{2}da + d^{2}ab - (bc + ad)(ac + bd) = - ac(b - d)(c - a)$ Now from identity: $ ac[a^{2}bc + b^{2}cd + c^{2}da + d^{2}ab - (ab + cd)(ac + bd)]$ $ + bd[a^{2}bc + b^{2}cd + c^{2}da + d^{2}ab - (bc + ad)(ac + bd)]$$ = ac[bd(b - d)(c - a)] + bd[ - ac(b - d)(c - a)] = 0$ it follows that: $ a^{2}bc + b^{2}cd + c^{2}da + d^{2}ab - (ab + cd)(ac + bd)\leq 0$ or $ a^{2}bc + b^{2}cd + c^{2}da + d^{2}ab - (bc + ad)(ac + bd) \leq 0$. Now the final part of solution is almost obvious: If $ a^{2}bc + b^{2}cd + c^{2}da + d^{2}ab - (ab + cd)(ac + bd)\leq 0$ then: $ a^{2}bc + b^{2}cd + c^{2}da + d^{2}ab \leq (ab + cd)(ac + bd) \leq (\frac {ab + cd + ac + bd}{2})^{2}$ $ = (\frac {(a + d)(b + c)}{2})^{2}$ $ \leq \frac {1}{4}[(\frac {a + b + c + d}{2})^{2}]^{2}$ $ = 4$. Case $ a^{2}bc + b^{2}cd + c^{2}da + d^{2}ab - (bc + ad)(ac + bd) \leq 0$ is analogous. Finding equality case reduces to some simple systems of equations.

Hard inequality.

Zarif wrote: Hard inequality. You're very funny , you only say easy , hard inequality and revive old topics to repost solutions.

The inequality is equivalent to: $$(a+c)(b+d)(ac+bd)\leq 4+(ac+bd)^2$$Since $(a+c)+(b+d)=4$, then $(a+c)(b+d)\leq 4$, so $(a+c)(b+d)(ac+bd)\leq 4(ac+bd)$. Replacing $x=(ac+bd)$, it's obvious that $4x\leq x^2+4$.