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lú rùi vậy cũng sai :(
\(BDT\Leftrightarrow\sqrt{\dfrac{c}{b}.\dfrac{a-c}{a}}+\sqrt{\dfrac{c}{a}.\dfrac{b-c}{b}}\le1\)
Áp dụng BĐT AM-GM ta có:
\(VT\le\dfrac{\dfrac{c}{b}+\dfrac{a-c}{a}}{2}+\dfrac{\dfrac{c}{a}+\dfrac{b-c}{b}}{2}=1\)
\(\sqrt{\frac{\left(a+bc\right)\left(b+ac\right)}{c+ab}}=\sqrt{\frac{\left(a^2+ab+ac+bc\right)\left(b^2+bc+ba+ac\right)}{c^2+ca+cb+ab}}=\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(b+a\right)\left(b+c\right)}{\left(c+a\right)\left(c+b\right)}}=a+b\left(a,b,c>0;a+b+c=1\right)\)
Bạn làm tương tự nha
\(\Rightarrow P=a+b+c+a+b+c=2\left(a+b+c\right)=2\)
\(\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{ab}\)
\(\Leftrightarrow\left(\sqrt{c\left(a-c\right)}\right)^2+\left(\sqrt{c\left(b-c\right)}\right)\le\left(\sqrt{ab}\right)^2\)
\(\Leftrightarrow c\left(a-c\right)+c\left(b-c\right)\le ab\)
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
a) \(A=\left(\sqrt{a}+\sqrt{b}\right)^2\le\left(\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{b}\right)^2=2a+2b\le2\)
Vậy GTLN của A là 2 \(\Leftrightarrow\hept{\begin{cases}\sqrt{a}=\sqrt{b}\\a+b=1\end{cases}\Leftrightarrow a=b=\frac{1}{2}}\)
b) Ta có : \(\left(\sqrt{a}+\sqrt{b}\right)^4\le\left(\sqrt{a}+\sqrt{b}\right)^4+\left(\sqrt{a}-\sqrt{b}\right)^4=2\left(a^2+b^2+6ab\right)\)
Tương tự : \(\left(\sqrt{a}+\sqrt{c}\right)^4\le2\left(a^2+c^2+6ac\right)\)
\(\left(\sqrt{a}+\sqrt{d}\right)^4\le2\left(a^2+d^2+6ad\right)\)
\(\left(\sqrt{b}+\sqrt{c}\right)^4\le2\left(b^2+c^2+6bc\right)\)
\(\left(\sqrt{b}+\sqrt{d}\right)^4\le2\left(b^2+d^2+6bd\right)\)
\(\left(\sqrt{c}+\sqrt{d}\right)^4\le2\left(c^2+d^2+6cd\right)\)
Cộng các vế lại, ta được :
\(B\le6\left(a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bd+2cd+2bc\right)=6\left(a+b+c+d\right)^2\)
\(\Rightarrow B\le6\)
Vậy GTLN của B là 6 \(\Leftrightarrow\hept{\begin{cases}\sqrt{a}=\sqrt{b}=\sqrt{c}=\sqrt{d}\\a+b+c+d=1\end{cases}}\Leftrightarrow a=b=c=d=\frac{1}{4}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT^2=\left(\sqrt{\left(a+c\right)\left(b+c\right)}+\sqrt{\left(a-c\right)\left(b-c\right)}\right)^2\)
\(\le\left(a+c+a-c\right)\left(b+c+b-c\right)\)
\(=2a\cdot2b=4ab=VP^2\)
\(\Rightarrow VT\le VP\) *ĐPCM*