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sửa đề lại bạn nhé =) \(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}\)
đặt \(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}=k\Rightarrow\hept{\begin{cases}a=kA\\b=kB\end{cases}va\hept{\begin{cases}c=kC\\d=kD\end{cases}}}\)
theo đề bài ta có \(\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=\sqrt{kA^2}+\sqrt{kB^2}+\sqrt{kC^2}+\sqrt{kD^2}\)
=\(\sqrt{k}\left(A+B+C+D\right)\left(1\right)\)
ta lại có \(\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}=\sqrt{\left(kA+kB+kC+kD\right)\left(A+B+C+D\right)}\)
=\(\sqrt{k\left(A+B+C+D\right)\left(A+B+C+D\right)}=\sqrt{k\left(A+B+C+D\right)^2}=\sqrt{k}\left(A+B+C+D\right)\left(2\right)\)
(1),(2)=> \(\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}\)
\(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}=\frac{a+b+c+d}{A+B+C+D}\)
\(\Rightarrow A.a=\frac{A^2\left(a+b+c+d\right)}{A+B+C+D}\Rightarrow\sqrt{Aa}=\frac{A\sqrt{a+b+c+d}}{\sqrt{A+B+C+D}}\)
Tương tự ta có: \(\sqrt{Bb}=\frac{B\sqrt{a+b+c+d}}{\sqrt{A+B+C+D}}\); \(\sqrt{Cc}=\frac{C\sqrt{a+b+c+d}}{\sqrt{A+B+C+D}}\); \(\sqrt{Dd}=\frac{D\sqrt{a+b+c+d}}{\sqrt{A+B+C+D}}\)
Cộng vế với vế:
\(\sqrt{Aa}+\sqrt{Bb}+\sqrt{Cc}+\sqrt{Dd}=\frac{\sqrt{a+b+c+d}}{\sqrt{A+B+C+D}}\left(A+B+C+D\right)=\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}\)
Làm cách này chắt đuoc
Ap dung BDT Bun-nhi-a-cop-xki ta co:
\(\left(\sqrt{Aa}+\sqrt{Bb}+\sqrt{Cc}+\sqrt{Dd}\right)^2\le\left(A+B+C+D\right)\left(a+b+c+d\right)\)
\(\Rightarrow\sqrt{Aa}+\sqrt{Bb}+\sqrt{Cc}+\sqrt{Dd}\le\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}\)Dau '=' xay ra khi \(\frac{A}{a}=\frac{B}{b}=\frac{C}{c}=\frac{D}{d}\)hay \(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}\)
Ma theo gia thuyet cua de bai thi:
\(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}\)
Nen dang thuc tren ton tai voi \(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}\)
Theo tính chất dãy tỉ số bằng nhau, đặt:
\(\dfrac{a}{A}=\dfrac{b}{B}=\dfrac{c}{C}=\dfrac{d}{D}=\dfrac{a+b+c+d}{A+B+C+D}=k>0\)
\(\Rightarrow a=kA;b=kB;c=kC;d=kD;a+b+c+d=k\left(A+B+C+D\right)\)
Do đó:
\(\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=\sqrt{kA^2}+\sqrt{kB^2}+\sqrt{kC^2}+\sqrt{kD^2}\)
\(=\sqrt{k}\left(A+B+C+D\right)\) (1)
\(\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}=\sqrt{k\left(A+B+C+D\right)^2}=\sqrt{k}\left(A+B+C+D\right)\) (2)
Từ (1);(2) suy ra điều phải c/m
Xét: \(\sqrt{\frac{a}{b+c+d}}=\frac{\sqrt{a}}{\sqrt{b+c+d}}=\frac{a}{\sqrt{a\left(b+c+d\right)}}\)
\(\sqrt{\frac{b}{c+d+a}}=\frac{\sqrt{b}}{\sqrt{c+d+a}}=\frac{b}{\sqrt{b\left(c+d+a\right)}}\)
\(\sqrt{\frac{c}{d+a+b}}=\frac{\sqrt{c}}{\sqrt{d+a+b}}=\frac{c}{\sqrt{c\left(d+a+b\right)}}\)
\(\sqrt{\frac{d}{a+b+c}}=\frac{\sqrt{d}}{\sqrt{a+b+c}}=\frac{d}{\sqrt{d\left(a+b+c\right)}}\)
\(\Rightarrow VT=\frac{a}{\sqrt{a\left(b+c+d\right)}}+\frac{b}{\sqrt{b\left(c+d+a\right)}}+\frac{c}{\sqrt{c\left(d+a+b\right)}}+\frac{d}{\sqrt{d\left(a+b+c\right)}}\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{\begin{matrix}\sqrt{a\left(b+c+d\right)}\le\frac{a+b+c+d}{2}\\\sqrt{b\left(c+d+a\right)}\le\frac{a+b+c+d}{2}\\\sqrt{c\left(d+a+b\right)}\le\frac{a+b+c+d}{2}\\\sqrt{d\left(a+b+c\right)}\le\frac{a+b+c+d}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{a}{\sqrt{a\left(b+c+d\right)}}\ge\frac{2a}{a+b+c+d}\\\frac{b}{\sqrt{b\left(c+d+a\right)}}\ge\frac{2b}{a+b+c+d}\\\frac{c}{\sqrt{c\left(d+a+b\right)}}\ge\frac{2c}{a+b+c+d}\\\frac{d}{\sqrt{d\left(a+b+c\right)}}\ge\frac{2d}{a+b+c+d}\end{matrix}\right.\)
\(\Rightarrow VT\ge\frac{2a}{a+b+c+d}+\frac{2b}{a+b+c+d}+\frac{2c}{a+b+c+d}+\frac{2d}{a+b+c+d}\)
\(\Rightarrow VT\ge\frac{2\left(a+b+c+d\right)}{a+b+c+d}\)
\(\Rightarrow VT\ge2\)
\(\Rightarrow\frac{a}{\sqrt{a\left(b+c+d\right)}}+\frac{b}{\sqrt{b\left(c+d+a\right)}}+\frac{c}{\sqrt{c\left(d+a+b\right)}}+\frac{d}{\sqrt{d\left(a+b+c\right)}}\ge2\)
\(\Leftrightarrow\sqrt{\frac{a}{b+c+d}}+\sqrt{\frac{b}{c+d+a}}+\sqrt{\frac{c}{d+a+b}}+\sqrt{\frac{d}{a+b+c}}\ge2\) ( đpcm )
Lời giải:
Áp dụng bất đẳng thức AM-GM:
\(\frac{b+c+d}{a}=\frac{b+c+d}{a}.1\leq \left(\frac{\frac{b+c+d}{a}+1}{2}\right)^2=\left(\frac{b+c+d+a}{2a}\right)^2\)
\(\sqrt{\frac{a}{b+c+d}}\geq \frac{2a}{a+b+c+d}\). Tương tự với các phân thức còn lại:
\(\Rightarrow \text{VT}\geq \frac{2(a+b+c+d)}{a+b+c+d}=2\) (đpcm)
Đặt \(\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=\frac{d}{D}=k\)\(\left(k>0\right)\)\(\Rightarrow\)\(a=Ak;b=Bk;c=Ck;d=Dk\)
\(\Rightarrow\)\(\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=A\sqrt{k}+B\sqrt{k}+C\sqrt{k}+D\sqrt{k}\)
\(=\sqrt{k}\left(A+B+C+D\right)\)
\(\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}=\sqrt{\left(Ak+Bk+Ck+Dk\right)\left(A+B+C+D\right)}\)
\(=\sqrt{k}\left(A+B+C+D\right)\)
=> đpcm