a+b+c=5

=>a+b=5-c

ab+b=b(a+1)\(\le\left(\frac{b + a + 1}{2}\right)^2=\frac{(a + b + 1)^2}{4}\)

=>\(ab + b \le \frac{(5 - c + 1)^2}{4} = \frac{(6 - c)^2}{4}\)

a,b,c>0

=>ab+b>0

=>\(\frac{c^3}{ab + b} \ge \frac{c^3}{\frac{(6 - c)^2}{4}} = \frac{4c^3}{(6 - c)^2}\)

4a+4b=4(a+b)=4(5-c)=20-4c

=>P>=20-4c+\(\frac{4c^3}{\left(6-c\right)^2}\)

\(\frac{4c^3}{(6 - c)^2} = \frac{4c^3}{(6 - c)^2} + (6 - c) + (6 - c) - 2(6 - c)\)

Theo Cosi, ta sẽ có:

\(\frac{4c^3}{(6 - c)^2} + (6 - c) + (6 - c) \ge 3 \sqrt[3]{\frac{4c^3}{(6 - c)^2} \cdot (6 - c) \cdot (6 - c)} = 3 \sqrt[3]{4c^3} = 3c \sqrt[3]{4}\)

\(\frac{4c^3}{(6 - c)^2}-\frac{25}{12}c+\frac{125}{36}\)

\(=\frac{144c^3 - (75c - 125)(36 - 12c + c^2)}{36(6 - c)^2}\)

\(= \frac{144c^3 - (75c^3 - 900c^2 + 2700c - 125c^2 + 1500c - 4500)}{36(6 - c)^2}\)

\(= \frac{69c^3 + 1025c^2 - 4200c + 4500}{36(6 - c)^2}\)

\(= \frac{(3c - 5)^2 (23c + 180)}{36(6 - c)^2} \ge 0 \quad \forall c \in (0, 5)\)

=>\(\frac{4c^3}{(6 - c)^2} \ge \frac{25}{12}c - \frac{125}{36}\) ∀c∈(0;5)

Dấu '=' xảy ra khi 3c-5=0

=>c=5/3

Do đó, ta có: \(P \ge 20 - 4c + \frac{25}{12}c - \frac{125}{36}\)

=>\(P \ge 20 - \frac{23}{12}c - \frac{125}{36}\)

=>\(P \ge 20 - 4\left(\frac{5}{3}\right) + \frac{4\left(\frac{5}{3}\right)^3}{\left(6 - \frac{5}{3}\right)^2}\)

=>\(P \ge 20 - \frac{20}{3} + \frac{4 \cdot \frac{125}{27}}{\frac{169}{9}} = \frac{40}{3} + \frac{500}{507} = \frac{7260}{507} = \frac{2420}{169}\)

Vậy: GTNN của P là 2420/169 khi a=b=c=5/3

Chắc áp dụng được Cauchy-Schwarz

Ta có: \(\sqrt[3]{\left(a+b\right).\frac{2}{3}.\frac{2}{3}}\le\frac{a+b+\frac{4}{3}}{3}=\frac{a+b}{3}+\frac{4}{9}\)

Tương tự rồi cộng các vế của BĐT lại, ta được: \(\sqrt[3]{\frac{4}{9}}P\le\frac{2\left(a+b+c\right)}{3}+\frac{4}{3}=2\Rightarrow P\le\sqrt[3]{18}\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)

Ta có \(\sqrt{8a^2+56}=\sqrt{8\left(a^2+7\right)}=2\sqrt{2\left(a^2+ab+2bc+2ca\right)}\)

\(=2\sqrt{2\left(a+b\right)\left(a+2c\right)}\le2\left(a+b\right)+\left(a+2c\right)=3a+2b+2c\)

Tương tự \(\sqrt{8b^2+56}\le2a+3b+2c;\)\(\sqrt{4c^2+7}=\sqrt{\left(a+2c\right)\left(b+2c\right)}\le\frac{a+b+4c}{2}\)

Do vậy \(Q\ge\frac{11a+11b+12c}{3a+2b+2c+2a+3b+2c+\frac{a+b+4c}{2}}=2\)

Dấu "=" xảy ra khi và chỉ khi \(\left(a,b,c\right)=\left(1;1;\frac{3}{2}\right)\)

a) \(P=1957\)

b) \(S=19.\)

Ta có : \(x+y\left(2+3x\right)=3\Leftrightarrow y=\frac{3-x}{3x+2}\)  ( vì x > 0 ) 

Khi đó : \(x+y=x+\frac{3-x}{3x+2}=\frac{3x^2+x+3}{3x+2}=A\) 

Chứng minh được :  \(A\ge\frac{-3+2\sqrt{11}}{3}\) => ... 

Ta có:\(\sum\dfrac{a^2+6a+3}{a^2+a}=\sum\left(1+\dfrac{5a+3}{a^2+a}\right)=3+\sum\dfrac{5a+3}{a^2+a}\)

Có BĐT phụ: \(\dfrac{5a+3}{a^2+a}\ge-\dfrac{7}{2}a+\dfrac{15}{2}\)đúng vì nó tương đương \(\left(7a+6\right)\left(a-1\right)^2\ge0\left(true\right)\)

Áp dụng tương tự ta có:

\(VT\ge3-\dfrac{7}{2}\left(a+b+c\right)+\dfrac{15}{2}.3\ge3-\dfrac{21}{2}+\dfrac{45}{2}=15\)

Dấu = xảy ra khi a=b=c=1