Xét bđt sau :\(\left(a+b^3\right)\left(m+n\right)\ge\left(\sqrt{am}+\sqrt{b^3n}\right)^2\)(đúng theo bunhia nhé)

Chon \(m=a;n=\frac{1}{b}\)khi đó :

\(\left(a+b^3\right)\left(\frac{1}{a}+b\right)\ge\left(\sqrt{a.a}+\sqrt{b^3.\frac{1}{b}}\right)^2\)

\(< =>\left(a+b^3\right)\left(\frac{1}{a}+b\right)\ge\left(a+b\right)^2\)

\(< =>a+b^3\ge\frac{\left(a+b\right)^2}{\frac{1}{a}+b}=\frac{a\left(a+b\right)^2}{1+ab}\)

Suy ra \(\frac{1}{a+b^3}\le\frac{1+ab}{a\left(a+b\right)^2}\)(*)

Bằng cách chứng minh tương tự ta được :\(\frac{1}{a^3+b}\le\frac{1+ab}{b\left(a+b\right)^2}\)(**)

Từ (*) và (**) suy ra : \(\frac{1}{a+b^3}+\frac{1}{a^3+b}\le\frac{1+ab}{a\left(a+b\right)^2}+\frac{1+ab}{b\left(a+b\right)^2}\)

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

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

Khi đó bài toán trở thành tìm GTLN của biểu thức :

\(A\le S=\left(a+b\right)\left(\frac{1}{ab\left(a+b\right)}+\frac{1}{a+b}\right)-\frac{1}{ab}=\frac{a+b}{ab\left(a+b\right)}+\frac{a+b}{a+b}-\frac{1}{ab}\)

\(=\frac{1}{ab}+1-\frac{1}{ab}=1\)

Vậy \(A_{max}=1\)đạt được khi ...

chuyên KHTN 2017 ?

Áp dụng hằng đẳng thức mà làm 

Hàng đẳng thức nào

nhung hdt dang nho do ban

Bài 2:b) \(9=\left(\frac{1}{a^3}+1+1\right)+\left(\frac{1}{b^3}+1+1\right)+\left(\frac{1}{c^3}+1+1\right)\)

\(\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\therefore\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le3\)

Ta sẽ chứng minh \(P\le\frac{1}{48}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)

Ai có cách hay?

1/Đặt a=1/x,b=1/y,c=1/z ->x+y+z=1.

2a) \(VT=\frac{\left(\frac{1}{a^3}+\frac{1}{b^3}\right)\left(\frac{1}{a}+\frac{1}{b}\right)}{\frac{1}{a}+\frac{1}{b}}\ge\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)^2}{\frac{1}{a}+\frac{1}{b}}\)

\(=\frac{\left[\frac{\left(a^2+b^2\right)^2}{a^4b^4}\right]}{\frac{a+b}{ab}}=\frac{\left(a^2+b^2\right)^2}{a^3b^3\left(a+b\right)}\ge\frac{\left(a+b\right)^3}{4\left(ab\right)^3}\)

\(\ge\frac{\left(a+b\right)^3}{4\left[\frac{\left(a+b\right)^2}{4}\right]^3}=\frac{16}{\left(a+b\right)^3}\)

Thôi đành dồn về bậc dễ chịu hơn vậy :))
\(9=\frac{1}{a^3}+1+\frac{1}{a^3}+\frac{1}{b^3}+1+\frac{1}{b^3}+\frac{1}{c^3}+1+\frac{1}{c^3}\)

\(\ge\frac{3}{a^2}+\frac{3}{b^2}+\frac{3}{c^2}\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le3\)

Đến đây ta có đánh giá bằng 2 cách như sau:

Cách 1:

Theo Bunhiacopski ta dễ có:

\(\left[2a+\left(b+c\right)\right]^2\ge4\cdot2a\left(b+c\right)\Rightarrow\frac{1}{\left(2a+b+c\right)^2}\le\frac{1}{8a\left(b+c\right)}\)

\(\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{\left(b+c\right)^2}\right]\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{4bc}\right]\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{8}\left(\frac{1}{b^2}+\frac{1}{c^2}\right)\right]\)

Khi đó:

\(P\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{8b^2}+\frac{1}{8c^2}+\frac{1}{4b^2}+\frac{1}{8a^2}+\frac{1}{8c^2}+\frac{1}{4c^2}+\frac{1}{8a^2}+\frac{1}{8b^2}\right]=\frac{3}{16}\)

Cách 2:

Áp dụng liên tiếp BĐT phụ dạng \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\) ta dễ có rằng:

\(\frac{1}{\left(2a+b+c\right)^2}=\left(\frac{1}{2a+b+c}\right)^2=\frac{1}{16}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)^2=\frac{1}{16}\left[\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(a+c\right)^2}+\frac{2}{\left(a+b\right)\left(a+c\right)}\right]\)

\(\Rightarrow16P\le\frac{2}{\left(a+b\right)^2}+\frac{2}{\left(b+c\right)^2}+\frac{2}{\left(c+a\right)^2}+\frac{2}{\left(a+b\right)\left(b+c\right)}+\frac{2}{\left(b+c\right)\left(c+a\right)}+\frac{2}{\left(c+a\right)\left(a+b\right)}\)

\(\le\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(b+c\right)^2}+\frac{4}{\left(c+a\right)^2}\)

\(\le4\cdot\frac{1}{16}\left[\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2+\left(\frac{1}{c}+\frac{1}{a}\right)^2\right]\)

\(=\frac{1}{2}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(\le\frac{1}{2}\cdot\left(3+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\le3\)

\(\Rightarrow P\le\frac{3}{16}\)

Đẳng thức xảy ra tại a=b=c=1

Áp dụng BĐT Bunhiacopski ta có:

\(\left(a^3+b\right)\left(\frac{1}{a}+b\right)\ge\left(a+b\right)^2;\left(b^3+a\right)\left(\frac{1}{b}+a\right)\ge\left(a+b\right)^2\)

\(\Rightarrow\frac{a+b}{a^3+b}\le\frac{\frac{1}{a}+b}{a+b};\frac{a+b}{b^3+a}\le\frac{\frac{1}{b}+a}{a+b}\)

\(\Leftrightarrow M\le\frac{\frac{1}{a}+b}{a+b}+\frac{\frac{1}{b}+a}{a+b}-\frac{1}{ab}=\frac{\frac{1}{a}+\frac{1}{b}+a+b}{a+b}-\frac{1}{ab}\)

\(=\frac{ab\left(a+b\right)+a+b-\left(a+b\right)}{ab\left(a+b\right)}=1\)

Dấu "=" xảy ra tại a=b=1

Áp dụng Bất Đẳng Thức \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\forall x;y;z\inℝ\)ta có

\(\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)=9abc>0\Rightarrow ab+bc+ca\ge3\sqrt{abc}\)

Ta có \(\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\forall a;b;c>0\)

Thật vậy \(\left(1+a\right)\left(1+b\right)\left(1+c\right)=1+\left(a+b+c\right)+\left(ab+bc+ca\right)+abc\)

\(\ge1+3\sqrt[3]{abc}+3\sqrt[3]{\left(abc\right)^2}+abc=\left(1+\sqrt[3]{abc}\right)^3\)

Khi đó \(P\le\frac{2}{3\left(1+\sqrt{abc}\right)}+\frac{\sqrt[3]{abc}}{1+\sqrt[3]{abc}}+\frac{\sqrt{abc}}{6}\)

Đặt \(\sqrt[6]{abc}=t\Rightarrow\sqrt[3]{abc}=t^2,\sqrt{abc}=t^3\)

Vì a,b,c>0 nên 0<abc\(\le\left(\frac{a+b+c}{3}\right)^2=1\Rightarrow0< t\le1\)

Xét hàm số \(f\left(t\right)=\frac{2}{3\left(1+t^3\right)}+\frac{t^2}{1+t^2}+\frac{1}{6}t^3;t\in(0;1]\)

\(\Rightarrow f'\left(t\right)=\frac{2t\left(t-1\right)\left(t^5-1\right)}{\left(1+t^3\right)^2\left(1+t^2\right)^2}+\frac{1}{2}t^2>0\forall t\in(0;1]\)

Do hàm số đồng biến trên (0;1] nên \(f\left(t\right)< f\left(1\right)\Rightarrow P\le1\)

\(\Rightarrow\frac{2}{3+ab+bc+ca}+\frac{\sqrt{abc}}{6}+\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\le1\)

Dấu "=" xảy ra khi a=b=c=1