2192

 Problem 1.  For $a,b,c\geqslant 0;a+b+c=1.$ Prove: $$\dfrac{1}{27}+{\dfrac {4}{27}}\,q-\dfrac{2}{3}{q}^{2}+{\dfrac {2}{27}}\sqrt { \left( 7 +13{q}^{2}-16q \right)  \left( 1-3q \right) ^{3}}\geqslant a^4 b+b^4 c+c^4 a.$$

(tthnew)

 Generalization.

 For $a,b,c\geqslant 0;t=\text{constant} ,t\in \mathbb{R};a+b+c=1.$ Prove that:

$$a^4 b +b^4 c+c^4 a +tabc \leqslant\dfrac{1}{3}qt-\dfrac{2}{3}{q}^{2}-{\dfrac {2}{27}}t+{\dfrac {4}{27}}\,q+\dfrac{1}{27}+{ \dfrac {2}{27}}\sqrt { \left( 1-3q \right) ^{3} \left( 5qt-t+{t}^ {2}-16q+7+13{q}^{2} \right) }$$

(tthnew)

Remark. When $t=0,$ we receive the first one.

The text in Github.

Update. 

Solution for the Generalization.  (tthnew) 

Let $$\text{M}=\dfrac{1}{27}+\dfrac{1}{3}qt-{\dfrac {2}{27}}t-\dfrac{2}{3}{q}^{2}+{\dfrac {4}{27}}q+{ \dfrac {2}{27}}\sqrt { \left( 1-3\,q \right) ^{3} \left( 5qt-t+{t}^ {2}-16q+7+13{q}^{2} \right) }.$$

Need to prove:

$$ \text{M}-\dfrac{1}{2}{p}^{3}q+\dfrac{1}{2}{p}^{2}r+\dfrac{3}{2}p{q}^{2}-\dfrac{5}{2}\,qr-tr{p}^{2}\geqslant \dfrac{1}{2} \left( {p}^{2}-q \right)  \left( a-b \right)  \left( b-c \right)  \left( a-c \right)$$

• If $(a-b)(b-c)(a-c) \leqslant 0,$ it's obvious.
• Else, after squaring$,$ the inequality become: $$\dfrac{\left( 2\,rt+32\,qr-14\,r-26\,{q}^{2}r-10\,qtr-2\,r{t}^{2}-1-14 \,{q}^{2}-M+7\,q+5\,Mq+2\,Mt-qt+3\,{q}^{2}t+12\,{q}^{3} \right) ^{2} }{\left( \dfrac{1}{4} \left( 2t-1+5\,q \right) ^{2}+{\frac {27}{4}}\, \left(q-1 \right) ^{2} \right) }\geqslant 0$$

Done.