Từ $0<a\le b\le c;b+c\le5;a+b+c=6$ suy ra: $1\le a\le 2;1\le b\le\dfrac{5}{2};2\le c\le 4$Từ đó ta có:$(a-1)(a-2)\le0 \Leftrightarrow a^2\le 3a-2 \Rightarrow a^3\le3a^2-2a\le7a-6$$(b-1)(b-\dfrac{5}{2})\le 0 \Leftrightarrow b^2\le\dfrac{7}{2}b-\dfrac{5}{2} \Rightarrow b^3\le\dfrac{7}{2}b^2-\dfrac{5}{2}b\le\dfrac{39}{4}b-\dfrac{35}{4}$$(c-2)(c-4)\le0 \Leftrightarrow c^2\le6c-8 \Rightarrow c^3\le6c^2-8c\le28c-48$Cộng các BĐT trên ta được:$P\le7a+\dfrac{39}{4}b+28c-\dfrac{251}{4}$ $\le7(a+b+c)+\dfrac{11}{4}(b+c)+\dfrac{73}{4}c-\dfrac{251}{4}$ $\le7.6+\dfrac{11}{4}.5+\dfrac{73}{4}.4-\dfrac{251}{4}=66$$\max P=66 \Leftrightarrow a=b=1;c=4$
Từ $0<a\le b\le c;b+c\le5;a+b+c=6$ suy ra: $1\le a\le 2;1\le b\le\dfrac{5}{2};\dfrac{5}{2}\le c\le 4$.Từ đó ta có:$(a-1)(a-2)\le0 \Leftrightarrow a^2\le 3a-2 \Rightarrow a^3\le3a^2-2a\le7a-6$$(b-1)(b-\dfrac{5}{2})\le 0 \Leftrightarrow b^2\le\dfrac{7}{2}b-\dfrac{5}{2} \Rightarrow b^3\le\dfrac{7}{2}b^2-\dfrac{5}{2}b\le\dfrac{39}{4}b-\dfrac{35}{4}$$(c-\dfrac{5}{2})(c-4)\le0 \Leftrightarrow c^2\le\dfrac{13}{2}c-10 \Rightarrow c^3\le\dfrac{13}{2}c^2-10c\le\dfrac{129}{4}c-65$Cộng các BĐT trên ta được:$P\le7a+\dfrac{39}{4}b+\dfrac{129}{4}c-\dfrac{319}{4}$ $\le7(a+b+c)+\dfrac{11}{4}(b+c)+\dfrac{45}{2}c-\dfrac{319}{4}$ $\le7.6+\dfrac{11}{4}.5+\dfrac{45}{2}.4-\dfrac{319}{4}=66$$\max P=66 \Leftrightarrow a=b=1;c=4$
Từ $0<a\le b\le c;b+c\le5;a+b+c=6$ suy ra: $1\le a\le 2;1\le b\le\dfrac{5}{2};2\le c\le 4$Từ đó ta có:$(a-1)(a-2)\le0 \Leftrightarrow a^2\le 3a-2 \Rightarrow a^3\le3a^2-2a\le7a-6$$(b-1)(b-\dfrac{5}{2})\le 0 \Leftrightarrow b^2\le\dfrac{7}{2}b-\dfrac{5}{2} \Rightarrow b^3\le\dfrac{7}{2}b^2-\dfrac{5}{2}b\le\dfrac{39}{4}b-\dfrac{35}{4}$$(c-2)(c-4)\le0 \Leftrightarrow c^2\le
6c-
8 \Rightarrow c^3\le
6c^2-
8c\le2
8c-
48$Cộng các BĐT trên ta được:$P\le7a+\dfrac{39}{4}b+2
8c-\dfrac{
251}{4}$ $\le7(a+b+c)+\dfrac{11}{4}(b+c)+\dfrac{
73}{
4}c-\dfrac{
251}{4}$ $\le7.6+\dfrac{11}{4}.5+\dfrac{
73}{
4}.4-\dfrac{
251}{4}=66$$\max P=66 \Leftrightarrow a=b=1;c=4$