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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 \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Bất đẳng thức cần chứng minh tương đương với:
\(\dfrac{a}{a+3\sqrt{bc}}+\dfrac{b}{b+3\sqrt{ca}}+\dfrac{c}{c+3\sqrt{ab}}\)
Ta áp dụng bất đẳng thức Cô si dạng \(2\sqrt{xy}\le x+y\) cho các căn thức ở mẫu, khi đó ta được:
\(\dfrac{a}{a+3\sqrt{bc}}+\dfrac{b}{b+3\sqrt{ca}}+\dfrac{c}{c+3\sqrt{ab}}\ge\) với biểu thức
\(\dfrac{2a}{2a+3b+3c}+\dfrac{2b}{3a+2b+3c}+\dfrac{2c}{3a+3b+2c}\)
Khi đó ta cần chứng minh:
\(\dfrac{2a}{2a+3b+3c}+\dfrac{2b}{3a+2b+3c}+\dfrac{2c}{3a+3b+2c}\ge\dfrac{3}{4}\)
Đặt: \(\left\{{}\begin{matrix}x=2a+3b+3c\\y=3a+2b+3c\\z=3a+3b+2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a=\dfrac{1}{4}\left(3y+3z-5x\right)\\2b=\dfrac{1}{4}\left(3z+3x-5y\right)\\2c=\dfrac{1}{4}\left(3x+3y-5z\right)\end{matrix}\right.\)
Khi đó đẳng thức trên được viết lại thành:
\(\dfrac{3y+3z-5x}{4x}+\dfrac{3z+3x-5y}{4y}+\dfrac{3x+3y-5z}{4z}\ge\dfrac{3}{4}\)
Hay: \(3\left(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{x}{z}+\dfrac{z}{x}\right)-15\ge3\)
Bất đẳng thức cuối cùng luôn đúng theo bất đẳng thức Cô si.
Vậy bất đẳng thức được chứng minh. Đẳng thức xảy ra khi và chỉ khi \(a=b=c\)
Đặt \(x=\sqrt{a};y=\sqrt{b};z=\sqrt{c}\)
Khi đó bđt đã tro chở thành:
\(\dfrac{yz}{x^2+3yz}+\dfrac{zx}{y^2+3zx}+\dfrac{xy}{z^2+3xy}\le\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{yz}{x^2+3yz}+\dfrac{1}{3}-\dfrac{zx}{y^2+3zx}+\dfrac{1}{3}-\dfrac{xy}{z^2+3xy}\ge1-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{x^2}{x^2+3yz}+\dfrac{y^2}{y^2+3zx}+\dfrac{z^2}{z^2+3xy}\ge\dfrac{3}{4}\) (đpcm)
\(\dfrac{a}{a+2\sqrt{\left(a+bc\right)}}=\dfrac{a}{a+2\sqrt{a\left(a+b+c\right)+bc}}=\dfrac{a}{a+2\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(=\dfrac{a}{a+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\)
\(\le\dfrac{a}{5^2}\left(\dfrac{1}{a}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\right)\)
\(=\dfrac{a}{25}\left(\dfrac{1}{a}+\dfrac{8}{\sqrt{\left(a+b\right)\left(a+c\right)}}\right)=\dfrac{1}{25}+\dfrac{8}{25}.\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
Tương tự:
\(\dfrac{b}{b+2\sqrt{b+ac}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\)
\(\dfrac{c}{c+2\sqrt{c+ab}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)
Cộng vế:
\(P\le\dfrac{3}{25}+\dfrac{4}{25}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{15}{25}=\dfrac{3}{5}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)
BĐT trở thành: \(\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}+\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}+\dfrac{zx}{\sqrt{x^2+z^2+2y^2}}\le\dfrac{1}{2}\)
Ta có:
\(x^2+z^2+y^2+z^2\ge\dfrac{1}{2}\left(x+z\right)^2+\dfrac{1}{2}\left(y+z\right)^2\ge\left(x+z\right)\left(y+z\right)\)
\(\Rightarrow\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}\le\dfrac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)
Tương tự: \(\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}\le\dfrac{1}{2}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)\)
\(\dfrac{zx}{\sqrt{z^2+x^2+2y^2}}\le\dfrac{1}{2}\left(\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)\)
Cộng vế với vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{zx+yz}{x+y}+\dfrac{xy+zx}{y+z}+\dfrac{yz+xy}{z+x}\right)=\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)
a) Ta có:
\(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)^2}{9}\ge\dfrac{\left(ab+bc+ca\right)}{3}\)
\(\Leftrightarrow\dfrac{a+b+c}{3}\ge\sqrt{\dfrac{ab+bc+ca}{3}}\)
Đẳng thức xảy ra khi $a=b=c.$
b) BĐT \(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
Hay là \(2\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\),
đúng.
Đẳng thức xảy ra khi $a=b=c.$
c) \(\Leftrightarrow\dfrac{\left(x^2+2\right)^2}{x^2+1}\ge4\Leftrightarrow x^4+4x^2+4\ge4x^2+4\Leftrightarrow x^4\ge0\)
Đẳng thức xảy ra khi $x=0.$
d) Xét hiệu hai vế đi bạn.
Ta có:
\(\left(a^2+b+c\right)\left(1+b+c\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+b+c}}\le\dfrac{a\sqrt{1+b+c}}{a+b+c}\)
Tương tự: \(\dfrac{b}{\sqrt{b^2+a+c}}\le\dfrac{b\sqrt{1+c+a}}{a+b+c}\) ; \(\dfrac{c}{\sqrt{c^2+b+a}}\le\dfrac{c\sqrt{1+a+b}}{a+b+c}\)
Cộng vế:
\(P\le\dfrac{a\sqrt{1+b+c}+b\sqrt{1+c+a}+c\sqrt{1+a+b}}{a+b+c}\)
Lại có:
\(a\sqrt{1+b+c}+b\sqrt{1+c+a}+c\sqrt{1+a+b}\)
\(=\sqrt{a}.\sqrt{a+ab+ac}+\sqrt{b}.\sqrt{b+bc+ab}+\sqrt{c}.\sqrt{c+ac+bc}\)
\(\le\sqrt{\left(a+b+c\right)\left(a+b+c+2ab+2bc+2ca\right)}\)
\(\Rightarrow P\le\dfrac{\sqrt{\left(a+b+c\right)\left(a+b+c+2ab+bc+ca\right)}}{a+b+c}=\sqrt{\dfrac{a+b+c+2ab+2bc+2ca}{a+b+c}}\)
Do đó ta chỉ cần chứng minh:
\(\dfrac{a+b+c+2ab+2bc+2ca}{a+b+c}\le3\Leftrightarrow a+b+c\ge ab+bc+ca\)
Thật vậy:
\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)=\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)\ge\left(ab+bc+ca\right)^2\)
\(\Rightarrow a+b+c\ge ab+bc+ca\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\((a^2+\frac{1}{b^2})(1+4^2)\geq (a+\frac{4}{b})^2\Rightarrow \sqrt{a^2+\frac{1}{b^2}}\geq \frac{1}{\sqrt{17}}(a+\frac{4}{b})\)
Hoàn toàn tương tự với những cái còn lại và cộng theo vế suy ra:
$S\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c})$
$\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{36}{a+b+c})$ theo BĐT Cauchy-Schwarz.
Áp dụng BĐT AM-GM:
\(a+b+c+\frac{9}{4(a+b+c)}\geq 3\)
\(\frac{135}{4(a+b+c)}\geq \frac{135}{4.\frac{3}{2}}=\frac{45}{2}\)
\(\Rightarrow a+b+c+\frac{36}{a+b+c}\geq \frac{51}{2}\)
\(\Rightarrow S\geq \frac{3\sqrt{17}}{2}\)
Vậy $S_{\min}=\frac{3\sqrt{17}}{2}$
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\((a^2+\frac{1}{b^2})(1+4^2)\geq (a+\frac{4}{b})^2\Rightarrow \sqrt{a^2+\frac{1}{b^2}}\geq \frac{1}{\sqrt{17}}(a+\frac{4}{b})\)
Hoàn toàn tương tự với những cái còn lại và cộng theo vế suy ra:
$S\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c})$
$\geq \frac{1}{\sqrt{17}}(a+b+c+\frac{36}{a+b+c})$ theo BĐT Cauchy-Schwarz.
Áp dụng BĐT AM-GM:
\(a+b+c+\frac{9}{4(a+b+c)}\geq 3\)
\(\frac{135}{4(a+b+c)}\geq \frac{135}{4.\frac{3}{2}}=\frac{45}{2}\)
\(\Rightarrow a+b+c+\frac{36}{a+b+c}\geq \frac{51}{2}\)
\(\Rightarrow S\geq \frac{3\sqrt{17}}{2}\)
Vậy $S_{\min}=\frac{3\sqrt{17}}{2}$
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow3\ge ab+bc+ca\)
\(\Rightarrow\left\{{}\begin{matrix}3+a^2\ge\left(a+c\right)\left(a+b\right)\\3+b^2\ge\left(a+b\right)\left(b+c\right)\\3+c^2\ge\left(a+c\right)\left(b+c\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{bc}{\sqrt{3+a^2}}\le\dfrac{bc}{\sqrt{\left(a+c\right)\left(a+b\right)}}\\\dfrac{ca}{\sqrt{3+b^2}}\le\dfrac{ca}{\sqrt{\left(a+b\right)\left(b+c\right)}}\\\dfrac{ab}{\sqrt{3+c^2}}\le\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{bc}{\sqrt{\left(a+c\right)\left(a+b\right)}}+\dfrac{ca}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(\Leftrightarrow VT\le\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\) (1)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}\le\dfrac{\dfrac{bc}{a+c}+\dfrac{bc}{a+b}}{2}\\\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{ca}{a+b}+\dfrac{ca}{b+c}}{2}\\\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{\dfrac{ab}{a+c}+\dfrac{ab}{b+c}}{2}\end{matrix}\right.\)
\(\Rightarrow\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{\left(\dfrac{bc}{a+c}+\dfrac{ab}{a+c}\right)+\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}\right)+\left(\dfrac{ab}{b+c}+\dfrac{ca}{b+c}\right)}{2}\)
\(\Rightarrow\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{a+b+c}{2}=\dfrac{3}{2}\) (2)
Xét \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(\Leftrightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\) (3)
Từ (1) , (2) , (3)
\(\Rightarrow VT\le\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(\Leftrightarrow\dfrac{bc}{\sqrt{a^2+3}}+\dfrac{ca}{\sqrt{b^2+3}}+\dfrac{ab}{\sqrt{c^2+3}}\le\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\) (đpcm)
Dấu " = " xảy ra khi \(a=b=c=1\)