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\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+ac+bc}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)=\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\) Chứng minh tương tự ta được:
\(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{b+a}+\dfrac{b}{b+c}\right);\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+a}+\dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)=\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\left(1+1+1\right)=\dfrac{3}{2}\) Dấu = xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{\sqrt{3}}\)
\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
Tương tự: \(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\) ; \(\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)
Cộng vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)
Ta có: \(\sqrt{a+bc}=\sqrt{\dfrac{a^2+abc}{a}}=\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)}{a}}\)
thiết lập tương tự ,bất đẳng thức cần chứng minh tương đương:
\(\Leftrightarrow\sum\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)}{a}}\ge\sqrt{abc}+\sqrt{a}+\sqrt{b}+\sqrt{c}\)
\(\Leftrightarrow\sum\sqrt{bc\left(a+b\right)\left(a+c\right)}\ge abc+\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
\(\Leftrightarrow\sum\sqrt{\left(b^2+ab\right)\left(c^2+ac\right)}\ge abc+\sum a\sqrt{bc}\)
Điều này luôn đúng theo BĐT Bunyakovsky:
\(\sum\sqrt{\left(b^2+ab\right)\left(c^2+ac\right)}\ge\sum\left(bc+a\sqrt{bc}\right)=abc+\sum a\sqrt{bc}\)
Dấu = xảy ra khi a=b=c=3
1: \(\Leftrightarrow a\sqrt{a}+b\sqrt{b}>=\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\)
=>\(\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b-\sqrt{ab}\right)>=0\)
=>\(\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)^2>=0\)(luôn đúng)
a) Áp dụng bất đẳng thức Cauchy Shwarz dạng Engel, ta có:
\(A=\dfrac{x^2}{x-1}+\dfrac{y^2}{y-1}\)
\(\ge\dfrac{\left(x+y\right)^2}{x+y-2}\)
Đặt \(x+y=a\left(a>0\right)\)
\(\Rightarrow A\ge\dfrac{a^2}{a-2}\)
\(=\dfrac{8\left(a-2\right)+\left(a^2-8a+16\right)}{a-2}\)
\(=8+\dfrac{\left(a-4\right)^2}{a-2}\ge8\)
Dấu "=" xảy ra khi \(x=y=2\)
b) Áp dụng bất đẳng thức Cauchy Shwarz dạng Engel, ta có:
\(A=\dfrac{x}{\sqrt{y}-1}+\dfrac{y}{\sqrt{z}-1}+\dfrac{z}{\sqrt{x}-1}\)
\(\ge\dfrac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{\sqrt{x}+\sqrt{y}+\sqrt{z}-3}\)
Đặt \(\sqrt{x}+\sqrt{y}+\sqrt{z}=a\left(a>0\right)\)
\(\Rightarrow A\ge\dfrac{a^2}{a-3}\)
\(=\dfrac{12\left(a-3\right)+\left(a^2-12a+36\right)}{a-3}\)
\(=12+\dfrac{\left(a-6\right)^2}{a-3}\ge12\)
Dấu "=" xảy ra khi x = y = z = 2
\(\sqrt{\dfrac{a}{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)
Tương tự: \(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c}\) ; \(\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Cộng vế:
\(VT\ge\dfrac{2a+2b+2c}{a+b+c}=2\)
Dấu "=" ko xảy ra nên \(VT>2\)
Cho a>0,b>0,c>0. Chứng minh \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}\sqrt{\dfrac{c}{a+b}}\ge2\)
\(S=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\)
\(\sqrt{a^2+\dfrac{1}{b^2}}=\dfrac{1}{\sqrt{17}}\sqrt{\left(a^2+\dfrac{1}{b^2}\right)\left(1+4^2\right)}\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}\right)\left(1\right)\)\(\left(bunhia\right)\)
\(tương-tự\Rightarrow\sqrt{b^2+\dfrac{1}{c^2}}\ge\dfrac{1}{\sqrt{17}}\left(b+\dfrac{4}{c}\right)\left(2\right)\)
\(\sqrt{c^2+\dfrac{1}{a^2}}\ge\dfrac{1}{\sqrt{17}}\left(c+\dfrac{4}{a}\right)\left(3\right)\)
\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}+b+\dfrac{4}{c}+c+\dfrac{4}{a}\right)\)
\(\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left[16a+\dfrac{4}{a}+16b+\dfrac{4}{b}+16c+\dfrac{4}{c}-15\left(a+b+c\right)\right]\)
\(\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left[2\sqrt{16a.\dfrac{4}{a}}+2\sqrt{16b.\dfrac{4}{b}}+2\sqrt{16c.\dfrac{4}{c}}-15.\dfrac{3}{2}\right]\left(am-gm\right)\)
\(\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left(16+16+16-\dfrac{45}{2}\right)=\dfrac{3\sqrt{17}}{2}\)
\(\Rightarrow MinS=\dfrac{3\sqrt{17}}{2}\Leftrightarrow a=b=c=\dfrac{1}{2}\)