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\(B=2\Sigma_{sym}\sqrt{ab}+3\left(a+b+c+d\right)\le6\left(a+b+c+d\right)\le6\)
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 \(\Leftrightarrow\hept{\begin{cases}\sqrt{a}=\sqrt{b}=\sqrt{c}=\sqrt{d}\\a+b+c+d=1\end{cases}}\Leftrightarrow a=b=c=d=\frac{1}{4}\)
\(B=\left[\left(\sqrt{a}+\sqrt{b}\right)^4+\left(\sqrt{c}+\sqrt{d}\right)^4\right]+\left[\left(\sqrt{a}+\sqrt{c}\right)^4+\left(\sqrt{b}+\sqrt{d}\right)^4\right]+\)
\(\left[\left(\sqrt{a}+\sqrt{d}\right)^4+\left(\sqrt{b}+\sqrt{c}\right)^4\right]\)\(\ge\frac{\left(a+b+2\sqrt{ab}+c+d+2\sqrt{cd}\right)^2+\left(a+c+2\sqrt{ac}+b+d+2\sqrt{bd}\right)^2+\left(a+d+2\sqrt{ad}+b+c+2\sqrt{bc}\right)^2}{2}\)
\(\ge\frac{\left(3a+3b+3c+3d+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}+2\sqrt{ad}+2\sqrt{cd}+2\sqrt{bd}\right)^2}{6}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{b}+\sqrt{c}\right)^2+\left(\sqrt{c}+\sqrt{d}\right)^2+\left(\sqrt{a}+\sqrt{c}\right)^2+\left(\sqrt{a}+\sqrt{d}\right)^2+\left(\sqrt{b}+\sqrt{d}\right)^2}{6}\)
tiếp tục sử dụng như hỗi nãy ta có:
\(\ge\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\right)^2}{2}\)
Bài 1 :
Áp dụng bất đẳng thức Cauchy ta có :
\(\frac{\left(x-1\right)^2}{z}+\frac{z}{4}\ge2\sqrt{\frac{\left(x-1\right)^2}{z}\frac{z}{4}}=\left|x-1\right|=1-x\)
\(\frac{\left(y-1\right)^2}{x}+\frac{x}{4}\ge2\sqrt{\frac{\left(y-1\right)^2}{x}\frac{x}{4}}=\left|y-1\right|=1-y\)
\(\frac{\left(z-1\right)^2}{y}+\frac{y}{4}\ge2\sqrt{\frac{\left(z-1\right)^2}{y}\frac{y}{4}}=\left|z-1\right|=1-z\)
\(\Rightarrow\frac{\left(x-1\right)^2}{z}+\frac{z}{4}+\frac{\left(y-1\right)^2}{x}+\frac{x}{4}+\frac{\left(z-1\right)^2}{y}+\frac{y}{4}\ge1-x+1-y+1-z\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{z}+\frac{\left(y-1\right)^2}{x}+\frac{\left(z-1\right)^2}{y}\ge3-\left(x+y+z\right)-\frac{x+y+z}{4}=3-2-\frac{2}{4}=\frac{1}{2}\)
Vậy GTNN của \(A=\frac{1}{2}\Leftrightarrow x=y=z=\frac{2}{3}\)
- Theo BĐT Cauchy ta có:
\(\sqrt{a.1}\le\dfrac{a+1}{2}\)
\(\sqrt{b.1}\le\dfrac{b+1}{2}\)
\(\sqrt{c.1}\le\dfrac{c+1}{2}\)
\(\sqrt{ab}\le\dfrac{a+b}{2}\)
\(\sqrt{bc}\le\dfrac{b+c}{2}\)
\(\sqrt{ca}\le\dfrac{c+a}{2}\)
\(\Rightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\dfrac{3\left(a+b+c\right)+3}{2}=\dfrac{3.3+3}{2}=6\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Mà ta có: \(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=6\)
\(\Rightarrow a=b=c=1\)
\(M=\dfrac{a^{30}+b^4+c^{1975}}{a^{30}+b^4+c^{2023}}=\dfrac{1^{30}+1^4+1^{1975}}{1^{30}+1^4+1^{2023}}=1\)
Theo BĐT C-S:
\(S^2=\left(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\right)^2\)
\(\le\left(1+1+1+1\right)\left(a+b+c+d\right)\)
\(=4\cdot\left(a+b+c+d\right)=4\left(a+b+c+d=1\right)\)
\(\Rightarrow S^2\le4\Rightarrow S\le2\)
Đẳng thức xảy ra khi a=b=c=d=1/4
a)Áp dụng AM-GM có:
\(a\sqrt{b-1}\le a.\dfrac{b-1+1}{2}=\dfrac{ab}{2}\)
\(b\sqrt{a-1}\le b.\dfrac{a-1+1}{2}=\dfrac{ab}{2}\)
\(\Rightarrow a\sqrt{b-1}+b\sqrt{a-1}\le\dfrac{ab}{2}+\dfrac{ab}{2}\)
\(\Leftrightarrow a\sqrt{b-1}+b\sqrt{a-1}\le ab\)
Dấu "=" xảy ra khi a=b=2
b)Áp dụng bđt bunhiacopxki có:
\(\left(\sqrt{ac}+\sqrt{bd}\right)^2=\left(\sqrt{a}.\sqrt{c}+\sqrt{b}.\sqrt{d}\right)^2\)\(\le\left[\left(\sqrt{a}\right)^2+\left(\sqrt{b}\right)^2\right]\left[\left(\sqrt{c}\right)^2+\left(\sqrt{d}\right)^2\right]=\left(a+b\right)\left(c+d\right)\)
\(\Rightarrow\sqrt{ac}+\sqrt{bd}\le\sqrt{\left(a+b\right)\left(c+d\right)}\)
Dấu "=" xảy ra khi \(\dfrac{\sqrt{a}}{\sqrt{c}}=\dfrac{\sqrt{b}}{\sqrt{d}}\Leftrightarrow ad=bc\)
\(b,\) Áp dụng BĐT Bunhiacopski:
\(\left(a+b\right)\left(c+d\right)=\left[\left(\sqrt{a}\right)^2+\left(\sqrt{b}\right)^2\right]\left[\left(\sqrt{c}\right)^2+\left(\sqrt{d}\right)^2\right]\\ \ge\left(\sqrt{ac}+\sqrt{bd}\right)^2\)
Dấu \("="\Leftrightarrow ad=bc\)
Mình làm hơi tắt nhé !
a, \(\left(5\sqrt{18}-3\sqrt{18}+4\sqrt{2}\right):\sqrt{2}\)
= \(5\sqrt{18:2}-3\sqrt{18:2}+4\sqrt{2:2}=15-9+4=10\)
b, \(\left(\sqrt{\dfrac{a^2}{d}}+\sqrt{\dfrac{b^2}{d}}-\sqrt{d}\right):\sqrt{d}\)
= \(\left(\sqrt{\dfrac{a^2}{d}}+\sqrt{\dfrac{b^2}{d}}-\sqrt{d}\right).\dfrac{1}{\sqrt{d}}=\dfrac{\sqrt{a^2}}{\sqrt{d}.\sqrt{d}}+\dfrac{\sqrt{b^2}}{\sqrt{d}.\sqrt{d}}-\dfrac{\sqrt{d}}{\sqrt{d}}=\dfrac{a}{d}+\dfrac{b}{d}-1\) = \(\dfrac{a+b}{d}-1\)
\(\left(\sqrt{a}+\sqrt{b}\right)^4\le\left(\sqrt{a}+\sqrt{b}\right)^4+\left(\sqrt{a}-\sqrt{b}\right)^4=2a^2+2b^2+6ab\)
\(tt\Rightarrow B\le6\left(a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd\right)=6\left(a+b+c+d\right)^2\le6\)
\(dấu"='\Leftrightarrow a=b=c=d=\dfrac{1}{4}\)