1/h=1/2(1/a+1/b)=1/2a+1/2b=(a+b)/2ab

=>(a+b/)2ab-1/h=0

quy dong len ta co

(a+b)h/2abh-2ab/2abh=0=> (ah+bh-2ab)/2abh=0 =>ah+bh-2ab=0

                                                                       =>ah+bh-ab-ab=0

                                                                         =>a(h-b)-b(a-h)=0  

                                                                           =>a(h-b)=b(a-h)

                                                                              =>a/b=(a-h)(h-b)

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

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

\(\frac{1}{h}=\frac{1}{2}.\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{h}=\frac{1}{2}.\frac{a+b}{ab}\Rightarrow\frac{1}{h}=\frac{a+b}{2ab}\)

\(\Rightarrow2ab=h\left(a+b\right)\Rightarrow ab+ab=ha+hb\)

\(\Rightarrow ab-hb=ah-ab\)

\(\Rightarrow\left(a-h\right).b=\left(h-b\right).a\)

\(\Rightarrow\frac{a-h}{h-b}=\frac{a}{b}\) (đpcm)

2) Theo nguyên lí Dirichlet, trong ba số \(a^2-1;b^2-1;c^2-1\) có ít nhất hai số nằm cùng phía với 1.

Giả sử đó là a² - 1 và b² - 1. Khi đó \(\left(a^2-1\right)\left(b^2-1\right)\ge0\Leftrightarrow a^2b^2-a^2-b^2+1\ge0\)

\(\Rightarrow a^2b^2+3a^2+3b^2+9\ge4a^2+4b^2+8\)

\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\ge4\left(a^2+b^2+2\right)\)

\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\left(c^2+3\right)\ge4\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\) (2)

Mà \(4\left[\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\right]\ge4\left(a+b+c+1\right)^2\) (3)(Áp dụng Bunhicopxki và cái ngoặc vuông)

Từ (2) và (3) ta có đpcm.

Sai thì chịu

Xí quên bài 2 b:v

b) Không mất tính tổng quát, giả sử \(\left(a^2-\frac{1}{4}\right)\left(b^2-\frac{1}{4}\right)\ge0\)

Suy ra \(a^2b^2-\frac{1}{4}a^2-\frac{1}{4}b^2+\frac{1}{16}\ge0\)

\(\Rightarrow a^2b^2+a^2+b^2+1\ge\frac{5}{4}a^2+\frac{5}{4}b^2+\frac{15}{16}\)

Hay \(\left(a^2+1\right)\left(b^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{3}{4}\right)\)

Suy ra \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{1}{4}+\frac{1}{2}\right)\left(\frac{1}{4}+\frac{1}{4}+c^2+\frac{1}{2}\right)\)

\(\ge\frac{5}{4}\left(\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c+\frac{1}{2}\right)^2=\frac{5}{16}\left(a+b+c+1\right)^2\) (Bunhiacopxki) (đpcm)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{2}\)

Cách nữa cho bài 2:

2a) Ta có: \(4\left(a^2+1+2\right)\left(1+1+\frac{\left(b+c\right)^2}{2}\right)\ge4\left(a+b+c+1\right)^2\)

Hay \(4\left(a^2+3\right)\left(2+\frac{\left(b+c\right)^2}{2}\right)\ge4\left(a+b+c+1\right)^2=VP\)

Như vậy ta quy bài toán về chứng minh: \(\left(b^2+3\right)\left(c^2+3\right)\ge4\left(2+\frac{\left(b+c\right)^2}{2}\right)\)

\(\Leftrightarrow b^2c^2+b^2+c^2+1\ge4bc\Leftrightarrow\left(bc-1\right)^2+\left(b-c\right)^2\ge0\)(đúng)

Đẳng thức xảy ra khi a = b = c = 1

b) Áp dụng BĐT Bunhiacopxki:\(\left(a^2+\frac{1}{4}+\frac{1}{4}+\frac{1}{2}\right)\left(\frac{1}{4}+b^2+c^2+\frac{1}{2}\right)\ge\frac{1}{4}\left(a+b+c+1\right)^2\)

\(\Rightarrow\frac{5}{4}\left(a^2+1\right)\left(b^2+c^2+\frac{3}{4}\right)\ge\frac{5}{16}\left(a+b+c+1\right)^2\)

Từ đó ta có thể quy bài toán về chứng minh: \(\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{4}\left(b^2+c^2+\frac{3}{4}\right)\)

...

Bài 3:Sửa đề a, b, c >0

Có:  \(\frac{a^3}{b^2}+\frac{a^3}{b^2}+b\ge3\sqrt[3]{\frac{a^6}{b^3}}=\frac{3a^2}{b}\)

Tương tự: \(\frac{2b^3}{c^2}+c\ge\frac{3b^2}{c};\frac{2c^3}{a^2}+a\ge\frac{3c^2}{a}\)

Cộng theo vế 3 BĐT trên: \(2\left(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\right)+a+b+c\ge3\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\)

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

\(\ge2\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+a+b+c\)

Từ đó ta có đpcm.

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