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a) \(BĐT\Leftrightarrow\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{ab}\)
\(\Leftrightarrow\sqrt{\frac{c\left(a-c\right)}{ab}}+\sqrt{\frac{c\left(b-c\right)}{ab}}\le1\)
\(\Leftrightarrow\sqrt{\frac{c}{b}\left(1-\frac{c}{a}\right)}+\sqrt{\frac{c}{a}\left(1-\frac{c}{b}\right)}\le1\)
Áp dụng AM-GM:\(VT\le\frac{1}{2}\left(\frac{c}{b}+1-\frac{c}{a}+\frac{c}{a}+1-\frac{c}{b}\right)=1\left(đpcm\right)\)
Dấu = xảy ra khi (a+b).c=ab
b) \(2+b+c+2+b+c\ge2\sqrt{\left(b+1\right)\left(c+1\right)}+2+b+c=\left(\sqrt{1+b}+\sqrt{1+c}\right)^2\ge4\left(1+a\right)\)
\(\Leftrightarrow b+c\ge2a\)
cau a) dung cosi
\(\sqrt{c\left(a-c\right)}\le\frac{a-c+c}{2}\) ap dung cosi cho hai so c va a-c
tuong tu voi cac so khac
\(BT\le\frac{a-c+c}{2}+\frac{b-c+c}{2}-\frac{a+b}{2}\)(bt la VT cua de)
=> DPCM
b)
dung cosi nhu cau a
lam nhanh luon
\(\sqrt{1+b}\ge\frac{b+1+1}{2}\)
tuong tu
\(BT\ge\frac{b+2}{2}+\frac{c+2}{2}\ge a+2\)
<=> b+c>=2a
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\(\sqrt{1+b}+\sqrt{1+c}=2\sqrt{1+a}\) (1)
⇒ \(b+c+2+2\sqrt{\left(1+b\right)\left(1+c\right)}=4a+4\)
⇒ \(b+c=4a+2-2\sqrt{\left(1+b\right)\left(1+c\right)}\)
Từ (1) ta lại có: \(2\sqrt{1+a}=\sqrt{1+b}+\sqrt{1+c}\ge2\sqrt[4]{\left(1+b\right)\left(1+c\right)}\)
⇒ \(1+a\ge\sqrt{\left(1+b\right)\left(1+c\right)}\) ⇒
\(b+c=4a+2-2\sqrt{\left(1+b\right)\left(1+c\right)}\ge4a+2-2\left(1+a\right)=2a\)
Vậy \(b+c\ge2a\), "=" xảy ra khi \(b=c=a\)
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Áp dụng bđt bunhiacopxki ta có:
\(\left(\sqrt{b+1}+\sqrt{c+1}\right)^2\le\left(b+1+c+1\right)\left(1^2+1^2\right)\)
\(\Leftrightarrow2\left(b+c+2\right)\ge4\left(a+1\right)\)
\(\Leftrightarrow b+c+2\ge2a+2\)
\(\Leftrightarrow b+c\ge2a\)
Chứng minh điều ngược lại đúng tức là. Cho a,b,c>0 thỏa \(b+c=2a\) thì \(\sqrt{b+1}+\sqrt{c+1}\le2\sqrt{a+1}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT=\left(\sqrt{b+1}+\sqrt{c+1}\right)^2\)
\(\le\left(1+1\right)\left(b+1+c+1\right)\)
\(=2\left(b+c+2\right)\le4\left(a+1\right)=VP\)
\(\Rightarrow\left(\sqrt{b+1}+\sqrt{1+c}\right)^2\le4\left(a+1\right)\)
\(\Rightarrow\sqrt{b+1}+\sqrt{1+c}\le\sqrt{4\left(a+1\right)}=2\sqrt{a+1}\)
BĐT cuối đúng hay ta có ĐPCM
Chứng minh điều ngược lại đúng, tức là :Cho a,b,c>0 thỏa \(b+c=2a\) thì \(\sqrt{b+1}+\sqrt{c+1}\le2\sqrt{a+1}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT^2=\left(\sqrt{b+1}+\sqrt{c+1}\right)^2\)
\(\le\left(1+1\right)\left(b+1+c+1\right)\)
\(=2\left(b+c+2\right)=2\left(2a+2\right)\)
\(=4\left(a+1\right)=2^2\sqrt{\left(a+1\right)^2}=VP^2\)
Vì \(VT^2\le VP^2\Rightarrow VT\le VP\)
BĐT kia đúng nên ta có ĐPCM