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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
\(BĐT\Leftrightarrow a^2+b^2-2+\left(\frac{ab+1}{a+b}\right)^2\ge0\)
\(\Leftrightarrow a^2+2ab+b^2-2ab-2+\left(\frac{ab+1}{a+b}\right)^2\ge0\)
\(\Leftrightarrow\left(a+b\right)^2-2\left(ab+1\right)+\left(\frac{ab+1}{a+b}\right)^2\ge0\)
\(\Leftrightarrow\left(a+b-\frac{ab+1}{a+b}\right)^2\ge0\) luôn đúng
a) \(4\left(a+b\right)ab=3\left(a-b\right)^2+\left(a+b\right)^2\Leftrightarrow4\left(a+b\right)ab=4a^2+4b^2-4ab\Leftrightarrow\left(a+b\right)ab=a^2+b^2-ab\) (đúng)
=> đẳng thức được cm
b) nếu nghĩ ra thì tớ giải cho
Ta có:
\(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge2\)
\(\Leftrightarrow a^2+b^2+2ab+\left(\frac{ab+1}{a+b}\right)^2\ge2\left(ab+1\right)\)
\(\Leftrightarrow\left(a+b\right)^2-2\left(a+b\right).\frac{ab+1}{a+b}+\left(\frac{ab+1}{a+b}\right)^2\ge0\)
\(\Leftrightarrow\left(a+b-\frac{ab+1}{a+b}\right)^2\ge0\)
Bất đẳng thức cuối luôn đúng, ta ta biến đổi tương đương nên bất đẳng thức ban đầu cũng đúng.
Ta có đpcm.
\(\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}\)
\(=\dfrac{a^2+\left(a-c\right)^2+c^2+2\left(ab-ac-bc\right)}{b^2+\left(b-c\right)^2+c^2+2\left(ab-ac-bc\right)}\)
\(=\dfrac{a^2+a^2-2ac+c^2+c^2+2ab-2ac-2bc}{b^2+b^2-2bc+c^2+c^2+2ab-2ac-2bc}\)
\(=\dfrac{2a^2+2c^2-4ac+2ab-2bc}{2b^2+2c^2-4bc+2ab-2ac}\)
\(=\dfrac{\left(a-c\right)^2+b\left(a-c\right)}{\left(b-c\right)^2+a\left(b-c\right)}\)
\(=\dfrac{\left(a-c\right)\left(a-c+b\right)}{\left(b-c\right)\left(a-c+b\right)}=\dfrac{a-c}{b-c}\left(đpcm\right)\)
\(a^2+b^2+\left(\dfrac{ab+1}{a+b}\right)^2>hoặc=2\)
<=>\(a^2+b^2+\left(\dfrac{ab+1}{a+b}\right)^2-2>hoặc=0\)
<=>\(\left(a+b\right)^2+\left(\dfrac{ab+1}{a+b}\right)^2-2\left(ab+1\right)>hoặc=0\)
<=>\(\left(a+b-\dfrac{ab+1}{a+b}\right)^2>hoặc=0\)
(đpcm)
chúc bạn học tốt ^ ^