More generaly we prove that if $a_1,\cdots,a_n$ are positive real numbers with sum $s$ and such that $a_1a_2+\cdots+a_na_1=k^2$, then for $r=0,1,2,3$ we have: \[ s_n= {{a_1^r}\over{s-a_1}}+{{a_2^r}\over{s-a_2}}+\cdots+{{a_n^r}\over{s-a_n}}\geq {{(2k)^{r-1}}\over{ (n-1)n^{r-2}}}. \] Indeed, the function $x\mapsto {{x^r}\over{s-x}}$ is convex for $s>x\geq 0$ and $r=0,1,2,3$. Hence by Jensen' inequality we have \[ s_n\geq {{n(s/n)^r}\over{s-s/n}}={{s^{r-1}}\over{(n-1)n^{r-2}}}. \] On the other hand, we know that $(a_1+\cdots+a_n)^2\geq 4(a_1a_2+\cdots+a_na_1)=4k^2$ and thus \[ s_n\geq {{(2k)^{r-1}}\over{(n-1)n^{r-2}}}. \] The stated problem corresponds to: $n=4, r=3$ and $k=1$.

or to write: x^3/(y+z+w)=x^4/x(z+y+w) for all, then use cauchy-schwarz! we will arrive to a simple thing...

$(w-x)^2+(x-y)^2+(y-z)^2+(z-w)^2\geq0.$ Hence, $w^2+x^2+y^2+z^2\geq1.$ $\frac{a^3}{b}\geq2a^2-ab,$ where $a>0,b>0.$ Hence, $\sum\frac{(3w)^3}{x+y+z}\geq3(6\sum w^2-2\sum wx)\geq3(5-2+(w-y)^2+(x-z)^2)\geq9.$ Hence, $\sum\frac{w^3}{x+y+z}\geq\frac{1}{3}.$

Problem:Let $ a;b;c;d$ be positive real numbers satisfying $ ab+bc+ca+da=1$.Prove that: $ \sum\frac{a^{3}}{b+c+d}\geq\frac{1}{3}$ My solution By AM-GM we have: $ \frac{a^{3}}{b+c+d}{18}+\frac{a}{6}+\frac{1}{12}\geq\frac{2a}{3}$ Similar,we have $ \sum\frac{a^{3}}{b+c+d}\geq\frac{a+b+c+d}{3}-\frac{1}{3}$ Note that $ ab+bc+ca+da=(a+c)(b+d)$,so $ (a+b+c+d)^{2}\geq 4(ab+bc+cd+da)=4$=>$ a+b+c+d\geq 2$ Now,we have $ \sum\frac{a^{3}}{b+c+d}\geq\frac{1}{3}$ Done!

Other problem: Let $ a;b;c;d$ be positive real numbers.Prove that: $ \frac{a^{3}+b^{3}+c^{3}}{b+c+d}\geq\sum a^{2}$

Let let $ a$, $ b$, $ c$ and $ d$ positive reals such us $ ab + bc + cd + da = 1$. Show that $ {\frac {{a}^{3}}{b + c + d}} + {\frac {{b}^{3}}{c + d + a}} + {\frac {{c}^{3}} {a + b + c}} + {\frac {{d}^{3}} {d + c + a}}\ge1/3$ Approach by campos: by hölder we have that $ \left(\sum \frac {a^3}{b + c + d}\right) \left(\sum a(b + c + d)\right) \left(\sum 1\right)^2\geq \left(\sum a\right)^4 \\ \\ \Rightarrow \sum \frac {a^3}{b + c + d}\geq \frac {(a + b + c + d)^4}{16(\sum a(b + c + d))}$ but it's easy to verify from the condition that $ (a + b + c + d)^2\geq 4(a + c)(b + d) = 4$ and $ 3(a + b + c + d)^2\geq 4(2 + 2ac + 2bd) = 4((a + b + c + d)^2 - (a^2 + b^2 + c^2 + d^2)) \\ \Leftrightarrow 4(a^2 + b^2 + c^2 + d^2)\geq (a + b + c + d)^2$ which is true by cauchy-schwarz Approach by Sung-yoon Kim: cronecker wrote: Let $ a$, $ b$, $ c$ and $ d$ positive reals such us $ ab + bc + cd + da = 1$. Show that $ {\frac {{a}^{3}}{b + c + d}} + {\frac {{b}^{3}}{c + d + a}} + {\frac {{c}^{3}} {a + b + c}} + {\frac {{d}^{3}} {d + c + a}}\ge1/3$ I think the original form is like this : $ \sum {\frac {a^3}{b + c + d}} \geq \frac {1}{3}$. By Cauchy, $ (LHS)(a(b + c + d) + b(c + d + a) + c(d + a + b) + d(a + b + c)) \geq (a^2 + b^2 + c^2 + d^2)^2$. So $ (LHS)\geq {\frac {(a^2 + b^2 + c^2 + d^2)^2}{2(ab + ac + ad + bc + bd + cd)}}$. -(*) By AM-GM, $ a^2 + b^2 + c^2 + d^2 \geq 2(ab + cd), 2(ac + bd), 2(ad + bc)$. So $ (a^2 + b^2 + c^2 + d^2)\geq ab + bc + cd + da$ and $ 3(a^2 + b^2 + c^2 + d^2) \geq ab + ac + ad + bc + bd + cd$. Multiplying this, we get that the RHS of (*) is not less than 1/3.

Fermat -Euler wrote: Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$. Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$. Follows Holder Inequality we have $ LHS\ge\frac{(x+y+z+w)^3}{4\cdot3(x+y+z+w)}=\frac{(x+y+z+w)^2}{4\cdot3(xy+yz+zw+wx)}\ge\frac{1}{3}$ because $ (x-y+z-w)^2\ge 0$

chien than wrote: Other problem: Let $ a;b;c;d$ be positive real numbers.Prove that: $ \frac {a^{3} + b^{3} + c^{3}}{b + c + d}\geq\sum a^{2}$ Sorry I cannot solve it and I don't remember if it was already posted. chien than wrote: Problem:Let $ a;b;c;d$ be positive real numbers satisfying $ ab + bc + ca + da = 1$.Prove that: $ \sum\frac {a^{3}}{b + c + d}\geq\frac {1}{3}$ My solution By AM-GM we have: $ \frac {a^{3}}{b + c + d}{18} + \frac {a}{6} + \frac {1}{12}\geq\frac {2a}{3}$ How to correct AM-GM ?

manlio wrote: chien than wrote: Other problem: Let $ a;b;c;d$ be positive real numbers.Prove that: $ \frac {a^{3} + b^{3} + c^{3}}{b + c + d}\geq\sum a^{2}$ Sorry I cannot solve it and I don't remember if it was already posted. chien than wrote: Problem:Let $ a;b;c;d$ be positive real numbers satisfying $ ab + bc + ca + da = 1$.Prove that: $ \sum\frac {a^{3}}{b + c + d}\geq\frac {1}{3}$ My solution By AM-GM we have: $ \frac {a^{3}}{b + c + d}{18} + \frac {a}{6} + \frac {1}{12}\geq\frac {2a}{3}$ How to correct AM-GM ? Dear manlio, It should be \[ \sum \frac{a^3+b^3+c^3}{b+c+d} \ge \sum a^2\] This inequality is easy with AM-GM Inequality, since \[ \frac{a^3}{b+c+d} +\frac{a(b+c+d)}{9} \ge \frac{2a^2}{3}\] \[ \Rightarrow \sum \frac{a^3}{b+c+d} \ge \frac{2}{3}\sum a^2 -\frac{2}{9}(ab+ac+ad+bc+bd+cd) \ge \frac{2}{3}\sum a^2-\frac{1}{3} \sum a^2 =\frac{1}{3}\sum a^2\] Similarly, also by AM-GM, \[ \frac{b^3}{b+c+d} +\frac{b(b+c+d)}{9} \ge \frac{2}{3}b^2\] \[ \Rightarrow \sum \frac{b^3}{b+c+d} \ge \frac{2}{3}\sum a^2 -\frac{1}{9}\sum a^2 -\frac{1}{9} (a+c)(b+d) -\frac{2}{9}(ac+bd) \ge \frac{2}{3}\sum a^2 -\frac{1}{9}\sum a^2 -\frac{1}{9}\sum a^2 -\frac{1}{9}\sum a^2 =\frac{1}{3}\sum a^2\] Similarly, we also have \[ \sum \frac{c^3}{b+c+d} \ge \frac{1}{3}\sum a^2\] Adding up these inequalities, we have done.

Exist a proof which is used Chebysev ineq ,Can_hang 2007.but your solution is very nice

Thank you very can_hang2007 for your nice solution.

The given condition implies $1 = wx+xy+yz+zx \le w^2+x^2+y^2+z^2$. Then by Cauchy, \[ \sum_{\text{cyc}} \frac{w^4}{wx+wy+wz} \ge \frac{(w^2+x^2+y^2+z^2)}{2(wx+xy+yz+zx+wy+xz)} \ge \frac{w^2+x^2+y^2+z^2}{2(wx+xy+yz+zx+wy+xz)} \ge \frac{1}{3}. \]

$\sum{\frac{x^3}{y+z+w}}=\sum{\frac{x^4}{xy+xz+xw}} \ge \frac{(\sum{x^2})^2}{2\sum{xy}} (\text{Titu's Lemma}) \ge \frac{2\sum{x^2}-1}{2\sum{xy}} ((\sum{x^2}-1)^2 \ge 0 \Leftrightarrow (\sum{x^2})^2 \ge 2\sum{x^2}-1) \ge \frac{2\sum{x^2}-1}{3\sum{x^2}}$ Now $\frac{2\sum{x^2}-1}{3\sum{x^2}} \ge \frac{1}{3}$ $\Leftrightarrow \sum{x^2} \ge 1$ But this follows with power-mean inequality combined with AM-GM: $\frac{\sum{x^2}}{4} \ge (\frac{\sum{x}}{4})^2 \ge \frac{4(y+w)(x+z)}{16}=\frac{1}{4}$ from which the result follows...It is easy to see that equality holds if and only if $x=y=z=w=\frac{1}{2}$

By cyclic AM-GM's $x^2+y^2+z^2+w^2\ge 1$, and also $\sum_{\text{sym}} x^2\ge \tfrac{1}{3}\left( \sum _{\text{sym}} xy\right)$. Hence by Cauchy \[\tfrac{w^4}{1-(x+z-w)y}+\tfrac{x^4}{1-(y+w-x)z}+\tfrac{y^4}{1-(x+z-y)w}+\tfrac{z^4}{1-(y+w-z)x}\ge\]\[\ge\tfrac{(w^2+x^2+y^2+z^2)^2}{2+2yw+2xz}=\tfrac{\left( \sum_{\text{sym}} x^2\right) ^2}{\sum_{\text{sym}} xy}\ge \tfrac{\sum_{\text{sym}} x^2}{3}\ge \tfrac{1}{3}.\]

We use Titu's Lemma, or Fractional Cauchy-Schwarz. Forming fourth powers in the numerator and applying the Lemma, $$\sum_{\text{cyc}} \frac{w^4}{w(x+y+z)} \geq \frac{(w^2+x^2+y^2 + z^2)^2}{2(wx+wy+wz+xy+yz+zx)} \geq \frac{1}{3}$$ With the given $wx+xy+yz + zw = 1$, the denominator can be rearranged into $2(1+wy+xz)$. Therefore, it remains to show that $$\frac{(w^2 + x^2 + y^2 + z^2)^2}{2(1+wy+xz)} \geq \frac{1}{3}$$ It is sufficient to show that $3(w^2 + x^2 + y^2 + z^2) \geq 2(1+wy+xz)$. By AM-GM, we have $w^2 + y^2 \geq 2wy$ and $x^2 + z^2 \geq 2xz$, which yield $w^2 + y^2 + x^2 + z^2 \geq 2(xz + wy)$ when added together. Furthermore, by the well known fact that $a_1^2 + a_2^2 + a_3^2 + a_4^2 \geq a_1a_2 + a_2a_3 + a_3a_4 + a_4a_1$, we have that $w^2 + x^2 + y^2 + z^2 \geq wx + xy + yz + zx =1$. The conclusion follows. (P.S: Sorry if a similar proof was already posted)

I think this is different enough to warrant a post. Oh well. By T2 Lemma and the given we get to $\text{Expression}\ge \frac{(w^2+x^2+y^2+z^2)^2}{2(1+wy+xz)}$. If $wy+zx\le \frac{1}{2}$, then we have, by the rearrangement ineq $w^2+x^2+y^2+z^2\ge wx+xy+zy+zx =1$, $\frac{(w^2+x^2+y^2+z^2)^2}{2(1+wy+xz)}\ge \frac{1}{2+2(wy+xz)}\ge \frac13$. Otherwise, we have $u=wy+xz\ge \frac12$. By am-gm, $w^2+x^2+y^2+z^2\ge 2wy+2zx=2u$, so our inequality rewrites as $\frac{(w^2+x^2+y^2+z^2)^2}{2(1+wy+xz)}\ge \frac{(4u^2)}{2(1+u)}\ge\frac13\iff 6u^2-u-1\ge0\iff (3u+1)(2u-1)\ge 0$ which is true. Combining both cases, we're done.

Solved using the AM-GM inequality, English is not my native language $(w+x+y+z)^2 \geq 4(w+y)(x+z) = 4(wx+xy+yz+zw) = 4$ $\Rightarrow w+x+y+z \geq 2$ We want to prove $$\sum_{cycle} \frac{w^3}{x+y+z} \geq \frac{1}{3}$$By AM-GM we have $\frac{w^3}{x+y+z}+\frac{x+y+z}{a}+\frac{1}{b} \geq \frac{3}{\sqrt[3]{ab}}w$ $$\Rightarrow \sum_{cycle} \frac{w^3}{x+y+z} \geq \frac{3(w+x+y+z)}{\sqrt[3]{ab}}-\frac{3(w+x+y+z)}{a}-\frac{4}{b}$$$$\Rightarrow \sum_{cycle} \frac{w^3}{x+y+z} \geq 3(w+x+y+z)(\frac{1}{\sqrt[3]{ab}}-\frac{1}{a})-\frac{4}{b}$$We want $\frac{1}{\sqrt[3]{ab}}-\frac{1}{a}>0 \Rightarrow a^2>b$ $$\Rightarrow \sum_{cycle} \frac{w^3}{x+y+z} \geq \frac{6}{\sqrt[3]{ab}}-\frac{6}{a}-\frac{4}{b}=\frac{1}{3}$$Let $\frac{1}{3}=\omega-\alpha-\beta$ $\Rightarrow \omega=\frac{6}{\sqrt[3]{ab}}, \alpha=\frac{6}{a}, \beta=\frac{4}{b}$ $\Rightarrow ab=\frac{216}{\omega^{3}}, a=\frac{6}{\alpha}, b=\frac{4}{\beta}$ $\Rightarrow 9\alpha\beta=\omega^{3}$ So any number $\omega, \alpha$ and $\beta$ that satisfy$\begin{cases} \omega-\alpha-\beta=\frac{1}{3}\\ 9\alpha\beta=\omega^{3}\\ a^2>b\end{cases}$ will give us the numbers $x, y$ to prove the inequality For example $\begin{cases}\omega=1\\\alpha=\frac{1}{3}\\\beta=\frac{1}{3}\\\end{cases}$ wil give us $\begin{cases}a=18\\b=12\end{cases}$ and we can use that to prove the original inequality If we plug in any $a$ and $b$ that don't follow this "rule" we will get $$\sum_{cycle} \frac{w^3}{x+y+z} \geq \gamma, \gamma < \frac{1}{3}$$This took me really long to figure this out, but when I did I was really happy

Fermat -Euler wrote: Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$. Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$. https://artofproblemsolving.com/community/c6h548439p3183012 https://artofproblemsolving.com/community/c6h2152301p15883499 SXTX （7）2008

Attachments:

It is easy to use Titu's Lemma. The equation, assume S $S\geq \frac{(w^2+x^2+y^2+z^2)^2}{2(xw+yw+zw+xy+yz+xz)}$ Since $wx+yx+yz+zw=1, w^2+y^2+z^2+x^2\geq 1; 2xz+2yw\geq 1$ $(w^2+x^2+y^2+z^2)^2\geq (wx+xy+yz+zw)^2=1$ by C-S inequality so $S\geq \frac{1}{2*1+1}=\frac{1}{3}$

Let $a,b,c,d$ are non-negative reals such that $ a b+a c+a d+b c+bd+c d=1 $. Show that $$ \frac{a^3}{b+c+d}+\frac{b^3}{c+d+a}+\frac{c^3}{a+b+d}+\frac{d^3}{a+b+c}\geq \frac {2}{9}$$

By the Cauchy-Schwarz inequality, we have: \[\quad \sum_{\text{cyc}} \frac{a^3}{b+c+d} = \sum_{\text{cyc}} \frac{a^4}{ab+ac+ad} \geq \frac{a^2 + b^2 + c^2 + d^2}{\sum_{\text{cyc}} (ab + ac + ad)} \geq \frac{1}{3}\]and we are done

Use $a$, $b$, $c$, and $d$ instead of $w$, $x$, $y$, and $z$, because I copied this from OTIS. Let $s = a+b+c+d$. Claim:$$\sum \frac{a^3}{s-a}\ge \frac{s^2}{12}.$$ Since $f(x)=\frac{x^3}{s-x}$ is convex, Jensen's inequality gives $$\frac{f(a)+f(b)+f(c)+f(d)}{4}\ge f(\frac{s}{4}).$$This is equivalent to $$f(a)+f(b)+f(c)+f(d)\ge \frac{s^2}{12}\ge \frac{1}{3}.$$The last part is simple; if we define $x = a+c$ and $y = b+d$, then $xy = 1$ so $x+y=s\ge 2.$