ko biết mới học lớp 6 thui à

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

\(\ge\frac{4}{a^2+b^2+2ab}+2.\sqrt{\frac{4ab}{4ab}}+\frac{1}{\left(a+b\right)^2}=4+2+1=7\)

Có : (a-b)^2>=0

<=>a^2+b^2>=2ab       (2)

<=>a^2+b^2+2ab>=4ab

<=>(a+b)^2>=4ab (1) hay 4ab<=(a+b)^2    (3)

Với a,b > 0 thì chia hai vế (1) cho ab.(a+b) ta được : a+b/ab >= 4/a+b <=> 1/a + 1/b >= 4/a+b     (4)

Áp dụng bđt (2) ; (3) và (4)  thì VT = (4/a^2+b^2 + 1/2ab) + (4ab+1/4ab)+1/4ab

>= 4/(a^2+b^2+2ab) + 2\(\sqrt{\frac{4ab.1}{4ab}}\)+ \(\frac{1}{\left(a+b\right)^2}\)

= 4/(a+b)^2 + 2 + 1/(a+b)^2 >= 4/1 + 2 + 1/1 = 7 => ĐPCM 

Dấu "=" xảy ra <=> a=b ; a+b=1 <=> a=b=1/2

a/ \(\frac{4bc-a^2}{bc+2a^2}.\frac{4ab-c^2}{ab+2c^2}.\frac{4ac-b^2}{ac+2b^2}\)

\(=\frac{4bc-\left(b+c\right)^2}{bc+2\left(b+c\right)^2}.\frac{4\left(-b-c\right)b-c^2}{\left(-b-c\right)b+2c^2}.\frac{4\left(-b-c\right)c-b^2}{\left(-b-c\right)c+2b^2}\)

\(=\frac{-\left(b-c\right)^2}{\left(c+2b\right)\left(b+2c\right)}.\frac{-\left(c+2b\right)^2}{-\left(b-c\right)\left(b+2c\right)}.\frac{-\left(b+2c\right)^2}{\left(b-c\right)\left(c+2b\right)}=1\)

Áp dụng Bat đẳng thức C.B.S dạng Angel

Dấu bằng xảy ra khi a=2;b=1;c=1

a)\(B=\frac{1}{a^2+b^2}+\frac{1}{ab}+4ab=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{2ab}+8ab-4ab\)

Áp dụng BĐT AM-GM ta có:

\(B=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{2ab}+8ab-4ab\)

\(\ge\frac{4}{\left(a+b\right)^2}+2\sqrt{\frac{1}{2ab}\cdot8ab}-\left(a+b\right)^2=7\)

Dấu "=" xảy ra khi \(\begin{cases}a=b\\a+b=1\end{cases}\)\(\Rightarrow a=b=\frac{1}{2}\)

Vậy \(Min_B=7\) khi \(a=b=\frac{1}{2}\)

b)\(C\ge\frac{1}{1-3ab\left(a+b\right)}+\frac{4}{ab\left(a+b\right)}\)

\(\ge\frac{16}{1-3ab\left(a+b\right)+3ab\left(a+b\right)}+\frac{1}{\frac{\left(a+b\right)^3}{4}}\ge16+4=20\)

Dấu "=" xảy ra khi \(\begin{cases}a=b\\a+b=1\end{cases}\)\(\Rightarrow a=b=\frac{1}{2}\)

Vậy \(Min_C=20\) khi \(a=b=\frac{1}{2}\)

thanks

bài 1:

=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)

ta có \(a^2+bc\ge2a\sqrt{bc}\)

=> \(\frac{a}{a^2+bc}\le\frac{a}{2a\sqrt{bc}}=\frac{1}{2\sqrt{bc}}\)

vì \(\frac12\cdot\frac{1}{\sqrt{bc}}\le\frac12\frac{\left(\frac{1}{b}+\frac{1}{c}\right)}{2}=\frac14\left(\frac{1}{b}+\frac{1}{c}\right)\)

=> \(\frac{a}{a^2+bc}\le\frac14\left(\frac{1}{b}+\frac{1}{c}\right)\)

=> \(\frac{a}{a^2+bc}+\frac{b}{b^2+ac}+\frac{c}{c^2+ab}\le\frac14\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right)=\frac12\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac32\)

dấu bằng xảy ra khi a=b=c=1(đpcm)

bài 2:

dự đoán điểm rơi a=3 và b=1

<=> \(S=\left(a^2-2a+9\right)+\left(a+\frac{9}{a}-6\right)+\left(b+\frac{1}{b}-2\right)+4a+2b+8\)

\(S=\left(a-3\right)^2+\frac{a^2-6a+9}{a}+\frac{b^2-2b+1}{b}+2\left(2a+b\right)+8\)

\(S=\left(a-3\right)^2+\frac{\left(a-3\right)^2}{a}+\frac{\left(b-1\right)^2}{b}+2\left(2a+b\right)+8\)

=> \(S\ge0+0+0+17+8=25\)

dấu bằng xảy ra khi và chỉ khi a-3=0=>a=3

b-1=0=>b=1