I have not yet solved this problem but it may be helpful to realize that we can perform the following parametrization: $ d = \frac{2x}{y + z}, e = \frac{2y}{x + z}, f = \frac{2z}{x + y} $ for some positive real numbers $ x, y, z $

v_Enhance wrote: Let $a,b,c,d,e,f$ be positive real numbers. Given that $def+de+ef+fd=4$, show that \[ ((a+b)de+(b+c)ef+(c+a)fd)^2 \geq\ 12(abde+bcef+cafd). \]Proposed by Allen Liu $ \bigstar$ The condition $def+de+ef+fd=4$ make some people feel this inequality is difficult, but for the above inequality, we only need \[ de+ef+fd \geq 3.\] $ \bigstar$ Now, we will prove the following inequality: If $a,b,c,x,y,z$ are positive real numbers, then \[ \big[(a+b)z+(b+c)x+(c+a)y\big]^2 \geq 4(abz+bcx+cay)(x+y+z).\] Without loss of generality, assume that $ c=\max\{a,b,c\}.$ Rewrite the above inequality as \[{ (a-b)^2z^2+(b-c)^2x^2+(c-a)^2y^2 \geq 2(a-b)(b-c)zx+2(b-c)(c-a)xy+2(c-a)(a-b)yz,}\] or \[ \big[(a-b)z-(b-c)x-(c-a)y\big]^2 \geq 4(b-c)(c-a)yz.\] The last inequality is obviously true because \[ \big[(a-b)z-(b-c)x-(c-a)y\big]^2 \geq 0 \geq 4(b-c)(c-a)yz.\] Now, for the original inequality. Replace $ z=de,\ x=ef,\ y=fd,$ we obtain \[\big[(a+b)de+(b+c)ef+(c+a)fd\big]^2\geq\ 4(abde+bcef+cafd)(de+ef+fd).\] Since $d,e,f>0$ and $ def+de+ef+fd=4,$ it is easily to see that $ de+ef+fd \geq 3.$ Hence, \[ \big[(a+b)de+(b+c)ef+(c+a)fd \big]^2\geq 12(abde+bcef+cafd). \] The proof is completed. Equality holds for $a=b=c$ and $d=e=f=1.$

quykhtn-qa1 wrote: $ \bigstar$ Now, we will prove the following inequality: If $a,b,c,x,y,z$ are positive real numbers, then \[ \big[(a+b)z+(b+c)x+(c+a)y\big]^2 \geq 4(abz+bcx+cay)(x+y+z).\] See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3544438#p3544438

Wolstenholme wrote: I have not yet solved this problem but it may be helpful to realize that we can perform the following parametrization: $ d = \frac{2x}{y + z}, e = \frac{2y}{x + z}, f = \frac{2z}{x + y} $ for some positive real numbers $ x, y, z $ I would just give a hint! After doing this step, we could prove this inequality by using $ a^2+b^2+c^2 \ge 2abcosC+2bccosA+2cacosB $.