VicKmath7 wrote: Prove for all positive real m, n, p, q and t=(m+n+p+q)/2 that : $ m/(t+n+p+q)+n/(t+m+p+q)+p/(t+m+n+q)+q/(t+m+n+p)=>4/5 $ According to Cauchy-Schwarz and Maclaurin inequalities, we get: $$\sum_{cyc} \frac{m}{t+n+p+q} \geq \frac{4t^2}{2t^2+2(mn+mp+mq+np+nq+qp)} \geq \frac{4t^2}{2t^2+3t^2}=\frac{4}{5}$$

Any solution without Maclaurin?

By C-S $$\sum_{cyc} \frac{m}{t+n+p+q}=3t\sum_{cyc} \frac{1}{t+n+p+q}-4 \ge \frac{48t}{4t+6t}-4=\frac{4}{5}$$

Thank you

VicKmath7 wrote: Any solution without Maclaurin? What about Chebyshev inequality? $$\sum_{cyc} \frac{m}{t+n+p+q} \geq \frac{t}{2}\left(\sum_{cyc} \frac{1}{t+n+p+q}\right) \geq \frac{t}{2}\times \frac{16}{\sum_{cyc}(t+n+p+q)}=\frac{t}{2}\times \frac{16}{10t}=\frac{4}{5}$$

Yes, better

VicKmath7 wrote: Yes, better Thanks

Let's look at only one part of the inequality: m/t+n+p+q =>? 1/5 /*(t+n+p+q) m =>? 1/5(3/2n+3/2p+3/2q+1/2m) /*10 10m =>? 3n+3p+3q+m 9m =>? 3(n+p+q) 3m =>? n+p+q We do that for every part and we get 3m =>? n+p+q 3n =>? m+p+q 3p =>? m+n+q 3q =>? m+n+p If the sum of these for inequalities gives us something true, then the sum of these inequalities also will be true m/(t+n+p+q) => 1/5 n/(t+m+p+q) => 1/5 p/(t+m+n+q) => 1/5 q/(t+m+n+p) => 1/5 and our problem will be solved. So let's sum up the first for inequalities We get 3m + 3n + 3p + 3q => 3(m+n+p+q) which is always true