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Đặt A =\(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\)
Vì a + b \(\ne\)0 nên A luôn được xác định.
Giả sử \(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge2\)
\(\Leftrightarrow\frac{\left(a^2+b^2\right)\left(a+b\right)^2}{\left(a+b\right)^2}+\frac{\left(ab+1\right)^2}{\left(a+b\right)^2}-\frac{2\left(a+b\right)^2}{\left(a+b\right)^2}\ge0\)
\(\Leftrightarrow\left(a^2+b^2\right)\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)(vì a + b \(\ne\)0)
\(\Leftrightarrow[\left(a^2+2ab+b^2\right)-2ab]\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)
\(\Leftrightarrow[\left(a+b\right)^2-2ab]\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)
\(\Leftrightarrow\left(a+b\right)^4-2ab\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)
\(\Leftrightarrow\left(a+b\right)^4-\left[2ab\left(a+b\right)^2+2\left(a+b\right)^2\right]+\left(ab+1\right)^2\ge0\)
\(\Leftrightarrow\left[\left(a+b\right)^2\right]^2-2\left(a+b\right)^2\left(ab+1\right)+\left(ab+1\right)^2\ge0\)
\(\left[\left(a+b\right)^2-\left(ab+1\right)^2\right]^2\ge0\)(luôn đúng)
Dấu bằng xảy ra
\(\Leftrightarrow\hept{\begin{cases}a+b\ne0\\\Leftrightarrow a=b\end{cases}}\Leftrightarrow a=b\left(a,b\ne0\right)\)
Vậy \(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge\)2 với a, b là các số thỏa mãn a+b \(\ne\)0
Dấu bằng xảy ra
\(\Leftrightarrow\hept{\begin{cases}a=b\\a+b\ne0\end{cases}\Leftrightarrow a=b}\)(a,b \(\ne\)0)
Vậy \(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge2\) với a, b là các số thỏa mãn \(a+b\ne0\)
BĐT tương đương
\(a^2+b^2+\frac{a^2b^2+2ab+1}{\left(a+b\right)^2}\ge2\)
<=>\(\left(a+b\right)^2-2+\frac{1}{\left(a+b\right)^2}+\frac{a^2b^2}{\left(a+b\right)^2}+\frac{2ab}{\left(a+b\right)^2}-2ab\ge0\)
<=>\(\left(a+b\right)^2-2.\left(a+b\right).\frac{1}{a+b}+\frac{a^2b^2}{\left(a+b\right)^2}-2.\left(ab-\frac{ab}{\left(a+b\right)^2}\right)\ge0\)
<=>\(\left(a+b-\frac{1}{a+b}\right)^2+\frac{a^2b^2}{\left(a+b\right)^2}-2.\left(\frac{ab\left(a+b\right)^2-ab}{\left(a+b\right)^2}\right)\ge0\)
<=>\(\left(\frac{\left(a+b\right)^2-1}{a+b}\right)^2+\frac{a^2b^2}{\left(a+b\right)^2}-2.\left(\frac{ab\left[\left(a+b\right)^2-1\right]}{\left(a+b\right)\left(a+b\right)}\right)\ge0\)
<=>\(\left(\frac{\left(a+b\right)^2-1}{a+b}\right)^2+\frac{a^2b^2}{\left(a+b\right)^2}-2.\frac{\left(a+b\right)^2-1}{a+b}.\frac{ab}{a+b}\ge0\)
<=>\(\left(\frac{\left(a+b\right)^2-1}{a+b}-\frac{ab}{a+b}\right)^2\ge0\left(\text{luôn đúng}\right)\)
=> dpcm
Ta có: \(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge2\)
\(\Leftrightarrow\left(a^2+b^2\right)\left(a+b\right)^2+\left(ab+1\right)^2\ge2\left(a+b\right)^2\)
\(\Leftrightarrow\left(a+b\right)^2\left[\left(a+b\right)^2-2ab\right]-2\left(a+b\right)^2+\left(ab+1\right)^2\ge0\)
\(\Leftrightarrow\left(a+b\right)^4-2ab\left(a+b\right)^2-2\left(a+b\right)^2+\left(ab+1\right)^2\ge0\)
\(\Leftrightarrow\left[\left(a+b\right)^2-ab-1\right]^2\ge0\)(đúng)
\(\Leftrightarrow dpcm\)
⇔(a2+b2)(a+b)2+(ab+1)2≥2(a+b)2
⇔(a+b)2[(a+b)2−2ab]−2(a+b)2+(ab+1)2≥0
⇔(a+b)4−2ab(a+b)2−2(a+b)2+(ab+1)2≥0
⇔[(a+b)2−ab−1]2≥0(đúng)
k mình đi
a) \(4\left(a+b\right)ab=3\left(a-b\right)^2+\left(a+b\right)^2\Leftrightarrow4\left(a+b\right)ab=4a^2+4b^2-4ab\Leftrightarrow\left(a+b\right)ab=a^2+b^2-ab\) (đúng)
=> đẳng thức được cm
b) nếu nghĩ ra thì tớ giải cho
\(BĐT\Leftrightarrow a^2+b^2-2+\left(\frac{ab+1}{a+b}\right)^2\ge0\)
\(\Leftrightarrow a^2+2ab+b^2-2ab-2+\left(\frac{ab+1}{a+b}\right)^2\ge0\)
\(\Leftrightarrow\left(a+b\right)^2-2\left(ab+1\right)+\left(\frac{ab+1}{a+b}\right)^2\ge0\)
\(\Leftrightarrow\left(a+b-\frac{ab+1}{a+b}\right)^2\ge0\) luôn đúng
a) Ta có: \(\left(a-b\right)^2\ge0\)
=>\(a^2+b^2-2ab\ge0\left(đpcm\right)\)
b) \(\left(a+b\right)^2\ge0\)
=> \(a^2+b^2+2ab\ge0\)
<=> \(a^2+b^2\ge-2ab\)
<=> \(\dfrac{a^2+b^2}{2}\ge ab\) (đpcm)
c) ta có: \(\left(a+1\right)^2=a^2+2a+1\)
\(a\left(a+2\right)=a^2+2a\)
Vậy từ 2 điều trên => \(a\left(a+2\right)< \left(a+1\right)^2\)
d) \(m^2+n^2+2\ge2\left(m+n\right)\) (*)
<=>m2 - 2m +1 +n2 - 2n +1 \(\ge0\)
<=> \(\left(m-1\right)^2+\left(n-1\right)^2\ge0\) (1)
(1) đúng => (*) đúng
d) Bạn ấy giải rồi ,mình không giải nữa
e) Theo BĐT cauchy ta có: \(\dfrac{a^2+b^2}{2}\ge ab\Rightarrow\dfrac{a^2+b^2}{ab}\ge2\)
\(\Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}\ge2\Leftrightarrow\left(\dfrac{a}{b}+1\right)+\left(\dfrac{b}{a}+1\right)\ge4\)
\(\Leftrightarrow\dfrac{a+b}{b}+\dfrac{a+b}{a}\ge4\)
\(\Rightarrow\left(a+b\right)\left(\dfrac{1}{b}+\dfrac{1}{a}\right)\ge4\) (đpcm)
Vậy..........
Lời giải:
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow \frac{ab+bc+ac}{abc}=0\Leftrightarrow ab+bc+ac=0\)
\(\Leftrightarrow 2(ab+bc+ac)=0\)
Cộng cả hai vế với \(a^2+b^2+c^2\) thì:
\(a^2+b^2+c^2+2(ab+bc+ac)=a^2+b^2+c^2\)
\(\Leftrightarrow (a+b+c)^2=a^2+b^2+c^2\)
Do đó ta có đpcm.
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT=\left(a^2+4\right)\left(b^2+9\right)\)
\(\ge\left(\sqrt{a^2b^2}+\sqrt{4\cdot9}\right)^2=\left(ab+36\right)^2=VP\)
Xảy ra khi \(\dfrac{a^2}{4}=\dfrac{b^2}{9}\Rightarrow\dfrac{a}{2}=\dfrac{b}{3}\Rightarrow b=\dfrac{3a}{2}\)
Khi đó \(A=\dfrac{a^2-ab+b^2}{a^2+ab+b^2}=\dfrac{a^2-a\cdot\dfrac{3a}{2}+\left(\dfrac{3a}{2}\right)^2}{a^2+a\cdot\dfrac{3a}{2}+\left(\dfrac{3a}{2}\right)^2}=\dfrac{7}{19}\)
xin lỗi bn nhưng bn có thể giải bằng cách khác ko , mk chưa học BĐT Cauchy-Schwart
\(a^2+b^2+\left(\dfrac{ab+1}{a+b}\right)^2>hoặc=2\)
<=>\(a^2+b^2+\left(\dfrac{ab+1}{a+b}\right)^2-2>hoặc=0\)
<=>\(\left(a+b\right)^2+\left(\dfrac{ab+1}{a+b}\right)^2-2\left(ab+1\right)>hoặc=0\)
<=>\(\left(a+b-\dfrac{ab+1}{a+b}\right)^2>hoặc=0\)
(đpcm)
chúc bạn học tốt ^ ^