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a) Mình sửa lại đề bài của bạn chút : Cần chứng minh \(\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}-\sqrt{ab}\le0\)
Áp dụng bất đẳng thức Bunhiacopxki , ta có : \(\left[\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\right]^2=\left(\sqrt{c}.\sqrt{a-c}+\sqrt{b-c}.\sqrt{c}\right)^2\le\left(c+b-c\right)\left(a-c+c\right)\)
\(\Rightarrow\left[\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\right]^2\le ab\Rightarrow\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{ab}\)
\(\Leftrightarrow\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}-\sqrt{ab}\le0\)(đpcm)
b) Ta có : \(\sqrt{1+b}+\sqrt{1+c}=2\sqrt{1+a}\)
Áp dụng bất đẳng thức Bunhiacopxki , ta có : \(\left(2\sqrt{1+a}\right)^2=\left(1.\sqrt{1+b}+1.\sqrt{1+c}\right)^2\le\left(1^2+1^2\right)\left(1+b+1+c\right)\)
\(\Leftrightarrow4\left(1+a\right)\le2\left(b+c+2\right)\Leftrightarrow4+4a\le2\left(b+c\right)+4\Leftrightarrow b+c\ge2a\)(đpcm)
Áp dụng BĐT Cauchy- schwarz:
\(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}\)
\(\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}=\frac{9}{\left(a+b+c\right)^2}\)
\(\Rightarrow\frac{1}{a^2+b^2+c^2}+\frac{2009}{ab+bc+ca}\)\(=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}\)\(+\frac{1}{ab+bc+ca}\)
\(+\frac{2007}{ab+bc+ca}\ge\frac{9}{\left(a+b+c\right)^2}+\frac{2007}{\frac{\left(a+b+c\right)^2}{3}}\)
\(=\frac{6030}{\left(a+b+c\right)^2}\ge670\)
(Dấu "="\(\Leftrightarrow a=b=c=1\))
Theo BĐT Cauchy-Schwarz, ta có:
\((a + b + c)^2 \le (1^2 + 1^2 + 1^2)(a^2 + b^2 + c^2) = 3 \cdot 3 = 9\)
=>a+b+c<=3(1)
\(3(a^2b^2 + b^2c^2 + c^2a^2) = (a^2 + b^2 + c^2)(a^2b^2 + b^2c^2 + c^2a^2)\)
Theo BĐT Am-Gm, ta có:
\(a^2b^2 + b^2c^2 + c^2a^2 \le \frac{(ab + bc + ca)^2}{3}\)
=>\(a^2b^2+b^2c^2+c^2a^2\le\frac{9}{3}=3\)
=>\(3(a^2b^2 + b^2c^2 + c^2a^2) \le 3(a + b + c)\)
\(\implies a^2b^2 + b^2c^2 + c^2a^2 \le a + b + c \quad (2)\)
Theo bđt Cauchy-Schwarz, ta có:
\(\text{VT} = \frac{1}{1+a^2b^2} + \frac{1}{1+b^2c^2} + \frac{1}{1+c^2a^2} \ge \frac{(1 + 1 + 1)^2}{(1 + a^2b^2) + (1 + b^2c^2) + (1 + c^2a^2)}\)
=>\(VT\ge\frac{9}{3 + (a^2b^2 + b^2c^2 + c^2a^2)}\)
=>\(VT\ge\frac{9}{3+\left(a+b+c\right)}\)
=>\(\text{VT} \ge \frac{9}{3 + a + b + c} \ge \frac{9}{2(a + b + c)}\) (ĐPCM)
Ta có: \(4=\left(\sqrt{a}+1\right)\left(\sqrt{b}+1\right)=\sqrt{ab}+\sqrt{a}+\sqrt{b}+1\)
\(\le\frac{a+b}{2}+\frac{a+1}{2}+\frac{b+1}{2}+1\Rightarrow a+b\ge2\)
Do đó \(P=\frac{a^2}{b}+\frac{b^2}{a}\ge\frac{\left(a+b\right)^2}{a+b}=a+b\ge2\)
Dấu bằng xảy ra khi a = b = 1