# Inequality $\left(2-a-b\right)f\left(\frac{\left(2-a-b\right)^{2}-2dc}{2-a-b}\right)+af\left(a\right)+bf\left(b\right)\leq 16/25$

It's an inequality which refines the following question:

Let $$a,b,d\in[0,1]$$ and $$c\in[0.5,1]$$ such that $$a+b+c+d=2$$ and $$a\geq d\ge c\ge b$$ then it seems we have :

$$af\left(a\right)+b\left(b\right)+cf\left(c\right)+df\left(d\right)\leq \left(2-a-b\right)f\left(\frac{\left(2-a-b\right)^{2}-2dc}{2-a-b}\right)+af\left(a\right)+bf\left(b\right)\leq 16/25$$

where :

$$f\left(x\right)=\frac{x}{\left(x^{2}+1\right)^{2}}$$

The LHS is just weighted Jensen's inequality because for $$x\in[0,1]$$ :

$$f''(x)=\frac{12x\left(x^{2}-1\right)}{\left(x^{2}+1\right)^{4}}\leq 0$$

So how to show the RHS? Perhaps in rewriting :

$$a\left(-f\left(\frac{\left(2-a-b\right)^{2}-2dc}{2-a-b}\right)+f\left(a\right)\right)+b\left(-f\left(\frac{\left(2-a-b\right)^{2}-2dc}{2-a-b}\right)+f\left(b\right)\right)+2f\left(\frac{\left(2-a-b\right)^{2}-2dc}{2-a-b}\right)$$

With the constraint above and $$a,c,d\in[1/\sqrt{3},2/3]$$ it seems we have :

$$\left(2-b\right)f\left(\frac{\left(\left(2-a-b\right)^{2}-2\left(2-a-b-c\right)c\right)}{2-a-b}\right)+bf\left(b\right)-16/25\leq 0$$

And :

$$f\left(\frac{\left(\left(2-a-b\right)^{2}-2\left(2-a-b-c\right)c\right)}{2-a-b}\right)-f\left(a\right)\ge 0$$

If this is true, we have the missing case described by the user alet on the Art of Problem Solving forum down below:

https://artofproblemsolving.com/community/c6h505313p2838484

Hint :

Using again Jensen's we need to show and the constraint of the question :

$$\left(2-b\right)f\left(\frac{a^{2}+c^{2}+d^{2}}{a+c+d}\right)+bf\left(b\right)-\frac{16}{25}\leq 0$$

Or :

$$\left(2-b\right)f\left(\frac{a^{2}+c^{2}+d^{2}}{2-b}\right)+bf\left(b\right)-\frac{16}{25}\leq 0$$

Plugging $$x=a^{2}+c^{2}+d^{2},b=y$$ we need to show :

$$a(x,y)=\left(2-y\right)f\left(\frac{x}{2-y}\right)+yf\left(y\right)-\frac{16}{25}\leq 0$$

then for $$y\geq x\geq 0$$:

$$a\left(\left(\frac{1}{2}+\frac{\frac{2}{5}x}{x+y}\right)^{2},\frac{1}{2}-\frac{\frac{1}{2}x}{x+1}\right)=-4(9147310144x^{16}+118863280640x^{15}y+83582321152x^{15}+605368876800x^{14}y^{2}+948796805120x^{14}y+271226571079x^{14}+1622992832000x^{13}y^{3}+4542570854400x^{13}y^{2}+2941070129240x^{13}y+437842970346x^{13}+2563169520000x^{12}y^{4}+11814473216000x^{12}y^{3}+13843727171300x^{12}y^{2}+4744200863760x^{12}y+386480907669x^{12}+2479486400000x^{11}y^{5}+18360956800000x^{11}y^{4}+35816811737000x^{11}y^{3}+22647478106200x^{11}y^{2}+4361668709640x^{11}y+182893716540x^{11}+1449436000000x^{10}y^{6}+17601830400000x^{10}y^{5}+55623597626250x^{10}y^{4}+59460220758000x^{10}y^{3}+21820996304300x^{10}y^{2}+2297592482400x^{10}y+38215246041x^{10}+471400000000x^{9}y^{7}+10235104000000x^{9}y^{6}+53376845025000x^{9}y^{5}+93474192397500x^{9}y^{4}+59355367947000x^{9}y^{3}+12645051018000x^{9}y^{2}+642249417960x^{9}y+26030970x^{9}+65625000000x^{8}y^{8}+3318080000000x^{8}y^{7}+31086000062500x^{8}y^{6}+90564719750000x^{8}y^{5}+95678103998750x^{8}y^{4}+36580243140000x^{8}y^{3}+4270017332700x^{8}y^{2}+73541293200x^{8}y-731636550x^{8}+461000000000x^{7}y^{8}+10094359375000x^{7}y^{7}+53141340375000x^{7}y^{6}+94371351075000x^{7}y^{5}+61332924705000x^{7}y^{4}+13632127983000x^{7}y^{3}+758719359000x^{7}y^{2}+254907000x^{7}y+1404652734375x^{6}y^{8}+17358116250000x^{6}y^{7}+56100372187500x^{6}y^{6}+62100252900000x^{6}y^{5}+24165927813750x^{6}y^{4}+2824495110000x^{6}y^{3}+53714340000x^{6}y^{2}+2426641406250x^{5}y^{8}+18502998125000x^{5}y^{7}+37593038250000x^{5}y^{6}+25323797175000x^{5}y^{5}+5387967337500x^{5}y^{4}+252312975000x^{5}y^{3}+2605733203125x^{4}y^{8}+12561697500000x^{4}y^{7}+15681900937500x^{4}y^{6}+5884989750000x^{4}y^{5}+525668062500x^{4}y^{4}+1786123437500x^{3}y^{8}+5323295625000x^{3}y^{7}+3740259375000x^{3}y^{6}+600800625000x^{3}y^{5}+765534765625x^{2}y^{8}+1292006250000x^{2}y^{7}+392343750000x^{2}y^{6}+188097656250xy^{8}+137953125000xy^{7}+20332031250y^{8})/(25(4x^{2}+8x+5)^{2}(46561x^{6}+174580x^{5}y+73122x^{5}+252150x^{4}y^{2}+269160x^{4}y+29061x^{4}+164500x^{3}y^{3}+384300x^{3}y^{2}+104580x^{3}y+40625x^{2}y^{4}+249000x^{2}y^{3}+147150x^{2}y^{2}+61250xy^{4}+94500xy^{3}+23125y^{4})^{2})\leq 0$$

Hope there is better proof .

Let :

$$r_b(x)=\left(2-b\right)f\left(\frac{x}{2-b}\right)+bf\left(b\right)-\frac{16}{25}$$

Then for $$0 and $$0\leq x\leq 1$$ we have :

$$r''_b(x)=-\frac{12(b-2)^{4}x(b-x-2)(b+x-2)}{(b^{2}-4b+x^{2}+4)^{4}}\leq 0$$

So we have that the derivative is decreasing on that domain which is :

$$r'_b(x)=\frac{(b-2)^{4}(b^{2}-4b-3x^{2}+4)}{(b^{2}-4b+x^{2}+4)^{3}}$$

It's not hard to see that for $$a,c,d\in[0.5,1/\sqrt{3}]$$ :

$$r'_b\left(a^{2}+c^{2}+d^{2}\right)\geq 0$$

So using a tangent line :

$$r\left(a^{2}+d^{2}+c^{2}\right)\geq r(3/4)$$

So with the three chord lemma $$\exists h\in(0,0.5)$$:

$$\frac{r\left(a^{2}+d^{2}+c^{2}\right)-r(3/4)}{a^{2}+d^{2}+c^{2}-(3/4)}\leq \frac{r\left(3/4+h\right)-r(3/4)}{h}$$

So :

$$r\left(a^{2}+d^{2}+c^{2}\right)\leq \frac{\left(r\left(\frac{3}{4}+h\right)-r\left(\frac{3}{4}\right)\right)}{h}\left(a^{2}+d^{2}+c^{2}-\frac{3}{4}\right)+r\left(\frac{3}{4}\right)$$

Now we choose $$h=0.5-b$$ .

We need to show :

$$\left(\frac{\left(r\left(\frac{3}{4}+h\right)-r\left(\frac{3}{4}\right)\right)}{h}\left(0.5-b+\left(0.5-b\right)^{2}\right)+r\left(\frac{3}{4}\right)\right)\leq 0$$

Wich is easy to handle as polynomial with a computer .