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\(A=\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\)
\(\Rightarrow A^2=\left(\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\right)^2\)\(=\left(\sqrt{c}.\sqrt{a-c}+\sqrt{c}.\sqrt{b-c}\right)^2\)
\(\Rightarrow A^2\le\left(c+b-c\right)\left(c+ a-c\right)\left(\text{ Bunhiacopxki}\right)\)
\(\Rightarrow A^2\le ab\Leftrightarrow A\le\sqrt{ab}\left(đpcm\right)\)
\(\)
Đề đánh bị lỗi.
Áp dụng bất đẳng thức Bunhiacopski:
\(\sqrt{c.\left(a-c\right)}+\sqrt{c.\left(b-c\right)}\le\sqrt{\left[\sqrt{c}^2+\sqrt{\left(a-c\right)}^2\right]\left[\sqrt{c}^2+\sqrt{\left(b-c\right)}^2\right]}\)
\(=\sqrt{\left(c+a-c\right)\left(c+b-c\right)}=\sqrt{ab}\)
1) Áp dụng BĐT bun-hi-a-cốp-xki ta có:
\(\left(a+d\right)\left(b+c\right)\ge\left(\sqrt{ab}+\sqrt{cd}\right)^2\)
\(\Leftrightarrow\sqrt{\left(a+d\right)\left(b+c\right)}\ge\sqrt{ab}+\sqrt{cd}\)( vì a,b,c,d dương )
Dấu " = " xảy ra \(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)
Sửa đề \(a;b>c>0\)
Giả sử \(\sqrt{ab}\ge\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\)
\(\Leftrightarrow ab\ge c\left(a-c\right)+c\left(b-c\right)+2c\sqrt{\left(a-c\right)\left(b-c\right)}\)
\(\Leftrightarrow ab-ac+c^2-bc+c^2-2c\sqrt{\left(a-c\right)\left(b-c\right)}\ge0\)
\(\Leftrightarrow\left(a-c\right)\left(b-c\right)-2c\sqrt{\left(a-c\right)\left(b-c\right)}+c^2\ge0\)
\(\Leftrightarrow\left(\sqrt{\left(a-c\right)\left(b-c\right)}\right)^2-2c\sqrt{\left(a-c\right)\left(b-c\right)}+c^2\ge0\)
\(\Leftrightarrow\left(\sqrt{\left(a-c\right)\left(b-c\right)}-c\right)^2\ge0\)đúng với \(\forall a;b>c>0\)
Bđt Bu-nhia-cop-xki \(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\), đẳng thức xảy ra khi \(ay=bx\)
a.
\(\left(2x+3y\right)^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2\le\left(2+3\right)\left(2x^2+3y^2\right)=5^2\)
\(\Rightarrow-5\le2x+3y\le5\)
b.
\(\sqrt{a+c}.\sqrt{b+c}+\sqrt{a-c}.\sqrt{b-c}\le\sqrt{a+c+a-c}.\sqrt{b+c+b-c}\)
\(=\sqrt{2a}.\sqrt{2b}=2\sqrt{ab}\)
Dấu bằng xảy ra khi \(\frac{\sqrt{a+c}}{\sqrt{a-c}}=\frac{\sqrt{b+c}}{\sqrt{b-c}}\), hay \(a=b\)
Thử lại với a = b thì \(VT=2a=2\sqrt{ab}=VP>\sqrt{ab}\) nên đề đã ra sai vế phải của bđt.
c.
bđt \(\Leftrightarrow\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)
d.
bđt \(\Leftrightarrow\left(a+c\right)^2+\left(b+d\right)^2\le a^2+b^2+c^2+d^2+2\sqrt{a^2+b^2}\sqrt{c^2+d^2}\)
\(\Leftrightarrow ac+bd\le\sqrt{a^2+b^2}.\sqrt{c^2+d^2}\)
bđt trên luôn đúng vì theo bđt Bu-nhia-cop-xki, ta có:
\(\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge\sqrt{\left(ac+bd\right)^2}=\left|ac+bd\right|\ge ac+bd\)
\(\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{ab}\)
\(\Leftrightarrow\left(\sqrt{c\left(a-c\right)}\right)^2+\left(\sqrt{c\left(b-c\right)}\right)\le\left(\sqrt{ab}\right)^2\)
\(\Leftrightarrow c\left(a-c\right)+c\left(b-c\right)\le ab\)
Thấy: \(c\left(a-c+b-c\right)\)
\(\Leftrightarrow ac-\left(c^2-cb+c^2\right)\)
\(c< b\Rightarrow ac< ab\)
Do đó: \(ac-\left(c^2-cb+c^2\right)< ab\)
Vậy: \(\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{ab}\)
ta cần cm \(\left(\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\right)^2\le ab\)
mà theo bunhia \(\left(\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\right)^2\le\left(c+b-c\right)\left(c+a-c\right)=ab\)
Đặt \(\sqrt{c.\left(a-c\right)}+\sqrt{c.\left(b-c\right)}\) = A
Ta có A^2 = \(\left(\sqrt{\left(a-c\right).c}+\sqrt{c.\left(b-c\right)}\right)^2\)
Áp dụng bđt bunhiacopxki ta có A^2 <= \(\left(\sqrt{a-c}^2+\sqrt{c^2}\right).\left(\sqrt{c^2}+\sqrt{b-c^2}\right)\)
= (a-c+c).(c+b-c) = ab
<=> A<= \(\sqrt{ab}\)=> ĐPCM
Dấu"=" <=> a-c = c và c = b-c
<=> a=b=2c>0
Ta có bất đẳng thức bunhicopxki
\(\sqrt{ax}+\sqrt{by}\le\sqrt{\left(a+x\right)\left(b+y\right)}\)
Áp dụng vào ta có:
\(\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{\left(a-c+c\right)\left(b-c+c\right)}\le\sqrt{ab}\)
Dấu bằng xảy ra khi a-c = b-c