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Dấu "=" không xảy ra
\(ĐK:a,b,c>0\)
\(\left\{{}\begin{matrix}\sqrt{a}+\sqrt{b}+\sqrt{c}=2\\\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}=\dfrac{1}{\sqrt{abc}}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=4\\\sqrt{abc}\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\right)=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)=4\\\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=1\end{matrix}\right.\)
\(\Rightarrow a+b+c=2\Rightarrow a=2-b-c\)
\(b+c\ge4abc\)
\(\Leftrightarrow b+c-4abc\ge0\)
\(\Leftrightarrow b+c-4\left(2-b-c\right)bc\ge0\)
\(\Leftrightarrow\left(b-4bc+4bc^2\right)+\left(c-4bc+4cb^2\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{b}-2c\sqrt{b}\right)^2+\left(\sqrt{c}-2b\sqrt{c}\right)^2\ge0\)
Mà do \(a,b,c>0\) nên dấu bằng không xảy ra
\(\Rightarrow b+c>4abc\)
a) Ta có BĐT:
\(a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\ge\left(a+b\right)ab\)
\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b+c\right)\)
\(\Rightarrow\frac{1}{a^3+b^3+abc}\le\frac{1}{ab\left(a+b+c\right)}\)
Tương tự cho 2 bất đẳng thức còn lại rồi cộng theo vế:
\(VT\le\frac{1}{ab\left(a+b+c\right)}+\frac{1}{bc\left(a+b+c\right)}+\frac{1}{ca\left(a+b+c\right)}\)
\(=\frac{a+b+c}{abc\left(a+b+c\right)}=\frac{1}{abc}=VP\)
Khi \(a=b=c\)
bđt cần c/m tương đương với:
\(\left(\frac{b+c}{\sqrt{a}}+\sqrt{a}\right)+\left(\frac{a+c}{\sqrt{b}}+\sqrt{b}\right)+\left(\frac{a+b}{\sqrt{c}}+\sqrt{c}\right)\ge2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+3\\ \ \)\(\left(a+b+c\right)\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\ge2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+3\)
Mặt khác:
\(a+b+c\ge\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{3}\)
\(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\ge\frac{9}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
=> \(VT\ge3\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
Ta cần c/m:
\(3\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\ge2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)+3\)
<=> \(\sqrt{a}+\sqrt{b}+\sqrt{c}\ge3\sqrt[3]{\sqrt{abc}}=3\)(BĐt Cô-si)
xong rồi bạn nhé
Bài toán phụ: chứng minh \(\left(x+y\right)^2\ge4xy\) với \(x,y\in R\)
Giải: Ta có: \(\left(x+y\right)^2\ge4xy\)\(\Leftrightarrow\)\(x^2+2xy+y^2-4xy\ge0\)
\(\Leftrightarrow\)\(x^2-2xy+y^2\ge0\)\(\Leftrightarrow\)\(\left(x-y\right)^2\ge0\) (luôn đúng).
Vậy \(\left(x+y\right)^2\ge4xy\) dấu "=" xảy ra \(\Leftrightarrow\)\(x=y.\)
Theo đề ta có \(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}=\frac{1}{\sqrt{abc}}\)\(\Leftrightarrow\)\(\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{\sqrt{abc}}=\frac{1}{\sqrt{abc}}\)
Suy ra \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=1\)
Mặt khác \(\sqrt{a}+\sqrt{b}+\sqrt{c}=2\)\(\Leftrightarrow\)\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=4\)
\(\Leftrightarrow\)\(a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)=4\)\(\Leftrightarrow\)\(a+b+c+2=4\)\(\Leftrightarrow\)\(a+b+c=2\)
Theo bài toán phụ ta có: \(\left(a+b+c\right)^2=\left[a+\left(b+c\right)\right]^2\ge4a\left(b+c\right)\)
Mà \(a+b+c=2\)\(\Rightarrow\)\(4\ge4a\left(b+c\right)\)\(\Leftrightarrow\)\(1\ge a\left(b+c\right)\)\(\Leftrightarrow\)\(b+c\ge a\left(b+c\right)^2\)
Do \(\left(b+c\right)^2\ge4bc\) nên \(a\left(b+c\right)^2\ge4abc\) hay \(b+c\ge4abc\) (đpcm).
Dấu "=" xảy ra khi \(\hept{\begin{cases}a=b+c\\b=c\end{cases}}\)\(\Leftrightarrow\)\(b=c=\frac{1}{2},\) \(a=1\)