Chứng minh \(a+b+\frac{a+b}{ab}-\frac{29}{5}\ge0\) biết \(a,b>0\) thỏa mãn \(a+b-\frac{4}{5}\le0\)
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đặt a = 2x+y+z ; b = 2y+z+x ; c = 2z+x+y => a+b+c = 4x+4y+4z
=> a - (a+b+c)/4 = x => x = (3a-b-c)/4 ; tương tự y = (3b-c-a)/4 ; z = (3c-a-b)/4
thay vào vế trái ta có
P = (3a-b-c)/4a + (3b-c-a)/4b + (3c-a-b)/4c =
= 9/4 - (b/4a + c/4a + c/4b + a/4b + a/4c + b/4c)
= 9/4 - (1/4)(b/a+a/b + c/a+a/c + c/b+b/c)
Côsi cho từng cặp ta có: b/a+a/b ≥ 2 ; c/a+a/c ≥ 2 ; c/b+b/c ≥ 2
=> b/a+a/b + c/a+a/c + c/b+b/c ≥ 6
=> -(1/4)(b/a+a/b +c/a+a/c + c/b+b/c) ≤ -6/4 thay vào P ta có:
P ≤ 9/4 - 6/4 = 3/4 (đpcm) ; dấu "=" khi a = b = c hay x = y = z
cách này tuy biến đổi dài nhưng dễ hiểu)
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Cách khác:
P = x/(2x+y+z) -1 + y/(2y+z+x) -1 + z/(2z+x+y) - 1 + 3
= -(x+y+z)/(2x+y+z) -(x+y+z)/(2y+z+x) -(x+y+z)/(2z+x+y) + 3
= -(x+y+z).[1/(2x+y+z) + 1/(2y+z+x) + 1/(2z+x+y)] + 3
- - -
Côsi cho 3 số:
2x+y+z + 2y+z+x + 2z+x+y ≥ 3.³√(2x+y+z)(2y+z+x)(2z+x+y)
=> 4(x+y+z) ≥ 3.³√(2x+y+z)(2y+z+x)(2z+x+y) (1*)
Côsi cho 3 số:
1/(2x+y+z)+1/(2y+z+x)+1/(2z+x+y) ≥ 3³√1/(2x+y+z)(2y+z+x)(2z+x+y) (2*)
Lấy (1*) *(2*) ta có:
4(x+y+z)[1/(2x+y+z) + 1/(2y+z+x) + 1/(2z+x+y)] ≥ 9
=> -(x+y+z).[1/(2x+y+z) + 1/(2y+z+x) + 1/(2z+x+y)] ≤ -9/4
thay vào P ta có:
P ≤ -9/4 + 3 = 3/4 (đpcm) ; dấu "=" khi x = y = z
Bạn ơi vì sao lại nhân với 9/4 mình tưởng chỉ nhân với 3/4 thôi chứ nhỉ
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Lời giải:
a) Áp dụng BĐT Cô-si cho các số dương:
$a^3+\frac{1}{8}+\frac{1}{8}\geq \frac{3}{4}a$
$b^3+\frac{1}{8}+\frac{1}{8}\geq \frac{3}{4}b$
$\Rightarrow a^3+b^3+\frac{1}{2}\geq \frac{3}{4}(a+b)=\frac{3}{4}$
$\Rightarrow a^3+b^3\geq \frac{1}{4}$ (đpcm)
Dấu "=" xảy ra khi $a=b=\frac{1}{2}$
b) Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{a^3+b^3}+\frac{3}{ab}=\frac{1}{a^2-ab+b^2}+\frac{1}{ab}+\frac{1}{ab}+\frac{1}{ab}\geq \frac{(1+1+1+1)^2}{a^2-ab+b^2+ab+ab+ab}\)
\(=\frac{16}{(a+b)^2}=16\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=\frac{1}{2}$
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Áp dụng tính chất dãy tỉ số bằng nhau; ta được:
\(\frac{ab+ac}{2}=\frac{bc+ba}{3}=\frac{ca+bc}{4}=\frac{ab+ac+bc+ba-\left(ca+bc\right)}{2+3-4}=\frac{2ab}{1}\)
Tương tự; ta được: \(\frac{ab+ac}{2}=\frac{bc+ba}{3}=\frac{ca+bc}{4}=\frac{bc+ba+ca+bc-\left(ab+ac\right)}{3+4-2}=\frac{2bc}{5}\)
\(\frac{ab+ac}{2}=\frac{bc+ba}{3}=\frac{ca+cb}{4}=\frac{ab+ac-\left(bc+ba\right)+ca+cb}{2-3+4}=\frac{2ac}{3}\)
Từ các điều trên; ta được:
\(\frac{2ac}{3}=\frac{2ab}{1}=\frac{2bc}{5}\)
\(\Rightarrow\frac{10ac}{15}=\frac{30ab}{15}=\frac{6bc}{15}\)
\(\Rightarrow10ac=30ab=6bc\)
\(\Rightarrow10ac=30ab\Rightarrow b=\frac{c}{3}\Rightarrow\frac{b}{5}=\frac{c}{15}\left(1\right)\)
\(30ab=6bc\Rightarrow5a=c\Rightarrow a=\frac{c}{5}\Rightarrow\frac{a}{3}=\frac{c}{15}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{a}{3}=\frac{b}{5}=\frac{c}{15}\left(ĐPCM\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\frac{ab+ac}{2}=\frac{bc+ba}{3}=\frac{ca+bc}{4}=\frac{ab+ac+bc+ba-\left(ca+bc\right)}{2+3-4}=\frac{2ab}{1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Không mất tính tổng quát giả sử \(c=max\left\{a,b,c\right\}\)
\(\Rightarrow2c\ge a+b\)
\(\Rightarrow c\ge\frac{a+b}{2}\)
Từ giả thiết \(\Rightarrow a,b\le1\)
\(\Rightarrow ab\le1\)( *)
Đặt \(P=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}-\frac{5}{2}\)
\(=\frac{1}{a+b}+\frac{1}{b+\frac{1-ab}{a+b}}+\frac{1}{a+\frac{1-ab}{a+b}}-\frac{5}{2}\)
Đặt \(S=\frac{1}{a+b+\frac{1}{a+b}}+a+b+\frac{1}{a+b}-\frac{5}{2}\)
Xét hiệu \(P-S=\)\(\frac{1}{a+b}+\frac{1}{b+\frac{1-ab}{a+b}}+\frac{1}{a+\frac{1-ab}{a+b}}-\frac{5}{2}-\)\(-\frac{1}{a+b+\frac{1}{a+b}}-a-b-\frac{1}{a+b}+\frac{5}{2}\)
\(=\frac{1}{\frac{ab+b^2+1-ab}{a+b}}+\frac{1}{\frac{a^2+ab+1-ab}{a+b}}-\frac{1}{\frac{\left(a+\right)^2+1}{a+b}}-\left(a+b\right)\)
\(=\frac{a+b}{b^2+1}+\frac{a+b}{c^2+1}-\left(a+b\right)\left[1+\frac{1}{1+\left(a+b\right)^2}\right]\)
Ta sẽ chứng minh \(\frac{a+b}{b^2+1}+\frac{a+b}{c^2+1}-\left(a+b\right)\left[1+\frac{1}{1+\left(a+b\right)^2}\right]\ge0\)
\(\Leftrightarrow\frac{a+b}{b^2+1}+\frac{a+b}{c^2+1}\ge\left(a+b\right)\left[1+\frac{1}{1+\left(a+b\right)^2}\right]\)
\(\Leftrightarrow\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge1+\frac{1}{1+\left(a+b\right)^2}\)
\(\Leftrightarrow\frac{2+a^2+b^2}{\left(1+a^2\right)\left(1+b^2\right)}\ge\frac{2+\left(a+b\right)^2}{1+\left(a+b\right)^2}\)
\(\Rightarrow\left(2+b^2+a^2\right)\left[1+\left(a+b\right)^2\right]\ge\left[2+\left(a+b\right)^2\right]\left(1+a^2\right)\left(1+b^2\right)\)
\(\Leftrightarrow2+2\left(a+b\right)^2+\left(a+b\right)^2\left(a^2+b^2\right)+a^2+b^2\ge\left[2+\left(a+b\right)^2\right]\left(1+a^2+b^2+a^2b^2\right)\)
\(\Leftrightarrow2+2\left(a+b\right)^2+\left(a+b\right)^2\left(a^2+b^2\right)+a^2+b^2-2a^2b^2-\left(a+b\right)^2\left(a^2+b^2\right)-\left(a+b\right)^2a^2b^2\)\(-2-2\left(a^2+b^2\right)-\left(a+b^2\right)\ge0\)
\(\Leftrightarrow-2a^2b^2-\left(a+b\right)^2a^2b^2+a^2+b^2-\left(a+b\right)^2\ge0\)
\(\Leftrightarrow ab\left[ab\left(a+b\right)^2+2ab-2\right]\le0\)
\(\Leftrightarrow ab\left(a+b\right)^2+2ab-2\le0\)( do a,b \(\ge0\))
\(\Leftrightarrow ab\left(a+b\right)^2\le2\left(1-ab\right)\)
\(\Leftrightarrow ab\left(a+b\right)^2\le2c\left(a+b\right)\) (1)
Mà \(c\ge\frac{a+b}{2}\)
\(\Rightarrow2c\left(a+b\right)\ge\left(a+b\right)^2\)
Ta có: \(\left(a+b\right)^2\ge ab\left(a+b\right)^2\)
\(\Leftrightarrow\left(a+b\right)^2\left(1-ab\right)\ge0\)( đúng do (*) )
\(\Rightarrow\left(1\right)\)đúng
\(\Rightarrow P-S\ge0\)
\(\Rightarrow P\ge S\)
Ta phải chứng minh \(S\ge0\)
\(\Leftrightarrow\frac{1}{a+b+\frac{1}{a+b}}+a+b+\frac{1}{a+b}\ge\frac{5}{2}\)
\(\Leftrightarrow\frac{a+b}{1+\left(a+b\right)^2}+\frac{1+\left(a+b\right)^2}{a+b}\ge\frac{5}{2}\) (2)
Đặt \(x=\frac{1+\left(a+b\right)^2}{a+b}\)
Ta có: \(1+\left(a+b\right)^2\ge2\left(a+b\right)\)
\(\Leftrightarrow\left(a+b-1\right)^2\ge0\)( đúng )
\(\Rightarrow x=\frac{1+\left(a+b\right)^2}{a+b}\ge2\)
=> (2) có dạng \(x+\frac{1}{x}\ge\frac{5}{2}\)
\(\Leftrightarrow2x^2-5x+2\ge0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\ge0\)( đúng )
\(\Rightarrow S\ge0\)mà \(P\ge S\)
\(\Rightarrow P\ge0\)
\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{5}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a+b=1\\ab+bc+ca=1\\ab\left[ab\left(a+b\right)^2+2ab-2\right]=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a=c=1;b=0\\b=c=1;a=0\end{cases}}\)
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BĐT tương đương : \(\frac{a\left(a+c+b-3b\right)}{1+ab}+\frac{b\left(b+a+c-3c\right)}{a+bc}+\frac{c\left(c+b+a-3a\right)}{1+ca}\ge0\)
\(\Leftrightarrow\frac{3a\left(1-b\right)}{1+ab}+\frac{3b\left(1-c\right)}{1+bc}+\frac{3c\left(1-a\right)}{1+ca}\ge0\)
\(\Leftrightarrow\frac{a\left(1-b\right)}{1+ab}+\frac{b\left(1-c\right)}{1+bc}+\frac{c\left(1-a\right)}{1+ca}\ge0\)
\(\Leftrightarrow\frac{a\left(1-b\right)}{1+ab}+1+\frac{b\left(1-c\right)}{1+bc}+1+\frac{c\left(1-a\right)}{1+ca}\ge3\)
\(\Leftrightarrow\frac{a+1}{1+ab}+\frac{b+1}{1+bc}+\frac{c+1}{1+ca}\ge3\)
Áp dụng BĐT Cosi ta có: \(\frac{a+1}{1+ab}+\frac{b+1}{1+bc}+\frac{c+1}{1+ca}\ge3\sqrt[3]{\frac{a+1}{1+ab}\cdot\frac{b+1}{1+bc}\cdot\frac{c+1}{1+ca}}\)
Ta phải chứng minh: \(\sqrt[3]{\frac{a+1}{1+ab}\cdot\frac{b+1}{1+bc}\cdot\frac{c+1}{1+ca}}\ge1\)
\(\Leftrightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge\left(1+ab\right)\left(1+bc\right)\left(1+ca\right)\)
Thật vậy \(\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge\left(1+ab\right)\left(1+bc\right)\left(1+ca\right)\)
\(\Leftrightarrow abc+ab+bc+ca+a+b+c+1\ge a^2b^2c^2+abc\left(a+b+c\right)+ab+bc+ca+1\)
\(\Leftrightarrow3\ge a^2b^2c^2+2abc\) (*)
Từ a+b+c=3 => \(3\ge3\sqrt[3]{abc}\Leftrightarrow abc\le1\)
=> (*) đúng
Vậy \(\frac{a\left(a+c-2b\right)}{1+ab}+\frac{b\left(b+a-2c\right)}{1+bc}+\frac{c\left(c+b-2a\right)}{1+ca}\ge0\)
Đẳng thức xảy ra <=> a=b=c=1
Áp dụng BĐT Cô - si ta có :
\(\frac{4}{5}\ge a+b\ge2\sqrt{a.b}\Rightarrow\sqrt{ab}\le\frac{2}{5}\Leftrightarrow ab\le\frac{4}{25}\)
\(\left(a+b\right)+\frac{a+b}{ab}\)
\(=\left[\left(a+b\right)+\frac{4}{25}.\frac{a+b}{ab}\right]+\frac{21}{25}.\frac{a+b}{ab}\)
\(\ge2\sqrt{\left(a+b\right).\frac{4}{25}.\frac{a+b}{ab}}+\frac{21}{25}.\frac{2\sqrt{ab}}{ab}\)
\(=2.\frac{2}{5}.\frac{a+b}{\sqrt{ab}}+\frac{21}{25}.\frac{2}{\sqrt{ab}}\)
\(\ge2.\frac{2}{5}.\frac{2\sqrt{ab}}{\sqrt{ab}}+\frac{21}{25}.\frac{2}{\sqrt{\frac{4}{25}}}\)
\(=\frac{8}{5}+\frac{21}{5}=\frac{29}{5}\)
Dấu ' = ' xảy ra khi và chỉ khi \(a=b=\frac{2}{5}\)
là không biết