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

\(\le\) 2(a+b)+(a+2c) = 3a+2b+2c

tương tự \(\sqrt{8b^2+56}\)\(\le\) 2a+3b+2c

\(\sqrt{4c^2+7}\) =\(\sqrt{4c^2+ab+2ac+2bc}\)= \(\sqrt{\left(a+2c\right)\left(b+2c\right)}\)\(\le\)(a+b+4c)/2

mẫu số \(\le\)3a+2b+2c+2a+3b+2c+a/2+b/2+2c=(11a+11b+12c)/2

 \(\Rightarrow\)  Q\(\ge\) 2

dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}ab+2bc+2ca=7\\2\left(a+b\right)=a+2c=b+2c\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}a=b=1\\c=1,5\end{cases}}\)

Vây...

Ta có: \(\sqrt{8a^2+56}=\sqrt{8\left(a^2+7\right)}=\sqrt{8\left(a^2+ab+2ab+2ac\right)}=2\cdot\sqrt{2\left(a+b\right)\left(a+2c\right)}\)

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

Tương tự\(\hept{\begin{cases}\sqrt{8b^2+56}\le2a+3b+2c\\\sqrt{4c^2+7}=\sqrt{4c^2+ab+2ac+2bc}=\sqrt{\left(a+2c\right)\left(b+2c\right)}\le\frac{a+b+4c}{2}\end{cases}}\)

=> Q>2 

Dấu "=" <=> \(\hept{\begin{cases}a=b=1\\c=1,5\end{cases}}\)

\(\sqrt{8a^2+56}=\sqrt{8\left(a^2+7\right)}=\sqrt{8\left(a^2+ab+2bc+2ac\right)}\)\(=\sqrt{8\left(a+b\right)\left(a+2c\right)}=\sqrt{4\left(a+b\right).2\left(a+2c\right)}\)

Áp dụng BĐT AM-GM cho các số không âm:

\(\sqrt{8a^2+56}=\sqrt{4\left(a+b\right).2\left(a+2c\right)}\le\frac{4\left(a+b\right)+2\left(a+2c\right)}{2}\)

\(\Rightarrow\)\(\sqrt{8a^2+56}\)\(\le3a+2b+2c\)

Tương tự:

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

\(\Rightarrow\sqrt{8a^2+56}+\sqrt{8b^2+56}+\sqrt{4c^2+7}\le\frac{11a+11b+12c}{2}\)

\(\Rightarrow P\ge\frac{11a+11b+12c}{\frac{11a+11b+12c}{2}}=2\)

\(''=''\Leftrightarrow a=b=\frac{2c}{3}=1\)

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 \(\sqrt{1+8a^3}=\sqrt{\left(1+2a\right)\left(1-2a+4a^2\right)}\le\frac{1+2a+1-2a+4a^2}{2}=1+2a^2\)(BĐT AM-GM)

Tương tự cho \(\sqrt{1+8b^2};\sqrt{1+8c^2}\)ta được \(P\ge\frac{1}{1+2a^2}+\frac{1}{1+2b^2}+\frac{1}{1+2c^2}\)

Mặt khác \(\frac{1}{1+2a^2}=\frac{1}{1+2a^2}+\frac{1+2a^2}{9}-\frac{1+2a^2}{9}\ge2\sqrt{\frac{1}{1+2a^2}\cdot\frac{1+2a^2}{9}}-\frac{2}{9}a^2-\frac{1}{9}=\frac{5-2a^2}{9}\)

Khi đó: \(P\ge\frac{5-2a^2}{9}-\frac{5-2b^2}{9}-\frac{5-2c^2}{9}\) \(=\frac{15-2\left(a^2+b^2+c^2\right)}{9}=\frac{15-2\cdot3}{9}=1\)

Vậy Min P=1

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a^2+b^2+c^2=3\\1+2a=1-2a+4a^2\\\frac{1}{1+2a^2}=\frac{1+2a^2}{9}\end{cases}}\)và vai trò a,b,c như nhau hay (a,b,c)=(1,1,1)

Áp dụng BĐT Bunyakovsky, ta có:

\(a+b+c\le\sqrt{3(a^2+b^2+c^2)}=\sqrt{3.3}=3\)

Áp dụng BĐT Cauchy, ta có:

\(A=\sum{\dfrac{1}{\sqrt{1+8a^3}}}=\sum{\dfrac{1}{\sqrt{(2a+1)(4a^2-2a+1)}}} \\\ge\sum{\dfrac{1}{\dfrac{4a^2+2}{2}}}=\sum{\dfrac{1}{2a^2+1}} \)

Ta cần chứng minh: \(\dfrac{1}{2a^2+1}\ge\dfrac{-4}{9}a+\dfrac{7}{9} \\<=>\dfrac{8a^3-14a^2+4a+2}{9(2a^2+1)}\ge0 \\<=>\dfrac{2(a-1)^2(4a+1)}{9(2a^2+1)}\ge0 (luôn\ đúng\ với\ mọi\ a>0) \\->\sum{\dfrac{1}{2a^2+1}}\ge\dfrac{-4}{9}(a+b+c)+\dfrac{21}{9}\ge\dfrac{-4}{9}.3+\dfrac{21}{9}=1 \\->A\ge1 \)

Đẳng thức xảy ra khi a = b = c = 1.

Vậy GTNN của A là 1 (khi a = b = c = 1).