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Đặt vế trái là T, ta có:
\(\dfrac{a}{\sqrt{b+1}}=\dfrac{a\sqrt{2}}{\sqrt{2}.\sqrt{b+1}}\ge\dfrac{a\sqrt{2}}{\dfrac{b+1+2}{2}}=\dfrac{a.2\sqrt{2}}{b+3}\)
Tương tự: \(\dfrac{b}{\sqrt{c+1}}\ge\dfrac{b.2\sqrt{2}}{c+3}\)
\(\dfrac{c}{\sqrt{a+1}}\ge\dfrac{c.2\sqrt{2}}{a+3}\)
Cộng vế theo vế các BĐT vừa chứng minh, ta được
\(T\ge2\sqrt{2}\left(\dfrac{a}{b+3}+\dfrac{b}{c+3}+\dfrac{c}{a+3}\right)=2\sqrt{2}\left(\dfrac{a^2}{ab+3a}+\dfrac{b^2}{bc+3b}+\dfrac{c^2}{ac+3c}\right)\)
\(T\ge2\sqrt{2}.\dfrac{\left(a+b+c\right)^2}{ab+bc+ca+3\left(a+b+c\right)}\)
\(T\ge2\sqrt{2}.\dfrac{\left(a+b+c\right)^2}{\dfrac{\left(a+b+c\right)^2}{3}+3\left(a+b+c\right)}\)
\(T\ge2\sqrt{2}.\dfrac{3^2}{\dfrac{3^2}{3}+9}=\dfrac{3\sqrt{2}}{2}\)(đpcm)
Đẳng thức xảy ra khi a=b=c=1
b) Đặt vế trái là N,ta có:
\(\sum\sqrt{\dfrac{a^3}{b+3}}=\sum\sqrt{\dfrac{a^4}{ab+3}}=\sum\dfrac{a^2}{\sqrt{ab+3}}=\sum\dfrac{2a^2}{\sqrt{4a\left(b+3\right)}}\ge\sum\dfrac{2a^2}{\dfrac{4a+b+3}{2}}=\sum\dfrac{4a^2}{4a+b+3}\)
\(\sum\dfrac{4a^2}{4a+b+3}\ge\dfrac{\left(2a+2b+2c\right)^2}{4a+b+3+4b+c+3+4c+a+3}=\dfrac{3}{2}\)(đpcm)
Đẳng thức xảy ra khi a=b=c=1
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 \(a=b=c=1\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 3\sqrt[3]{\frac{1}{(a+1)(b+1)(c+1)}}$
$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}$
Cộng theo vế và thu gọn:
$\frac{a+1}{a+1}+\frac{b+1}{b+1}+\frac{c+1}{c+1}\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$
$\Leftrightarrow 3\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$
$\Rightarrow (a+1)(b+1)(c+1)\geq (1+\sqrt[3]{abc})^3$
Ta có đpcm.
Bài 2:
$a^3+a^3+a^3+a^3+b^3+c^3\geq 6\sqrt[6]{a^{12}b^3c^3}=6a^2\sqrt{bc}$
$b^3+b^3+b^3+b^3+a^3+c^3\geq 6b^2\sqrt{ac}$
$c^3+c^3+c^3+c^3+a^3+b^3\geq 6c^2\sqrt{ab}$
Cộng theo vế và rút gọn thu được:
$a^3+b^3+c^3\geq a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}$
Ta có đpcm.
Dấu "=" xảy ra khi $a=b=c$
Nice proof, nhưng đã quy đồng là phải thế này :v
\(BDT\Leftrightarrow\left(2a-\sqrt{a^2+3}\right)+\left(2b-\sqrt{b^2+3}\right)+\left(2c-\sqrt{c^2+3}\right)\)
\(\Leftrightarrow\dfrac{a^2-1}{2a+\sqrt{a^2+3}}+\dfrac{b^2-1}{2b+\sqrt{b^2+3}}+\dfrac{c^2-1}{2c+\sqrt{c^2+3}}\ge0\)
\(\Leftrightarrow\dfrac{a^2-1}{2a+\sqrt{a^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{a}-a\right)+\dfrac{b^2-1}{2b+\sqrt{b^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{b}-b\right)+\dfrac{c^2-1}{2c+\sqrt{c^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{c}-c\right)\ge0\)
\(\Leftrightarrow\left(a^2-1\right)\left(\dfrac{1}{2a+\sqrt{a^2+3}}-\dfrac{1}{4a}\right)+\left(b^2-1\right)\left(\dfrac{1}{2b+\sqrt{b^2+3}}-\dfrac{1}{4b}\right)+\left(c^2-1\right)\left(\dfrac{1}{2c+\sqrt{a^2+3}}-\dfrac{1}{4c}\right)\ge0\)
\(\Leftrightarrow\dfrac{\left(a^2-1\right)\left(2a-\sqrt{a^2+3}\right)}{a\left(2a+\sqrt{a^2+3}\right)}+\dfrac{\left(b^2-1\right)\left(2b-\sqrt{b^2+3}\right)}{b\left(2b+\sqrt{b^2+3}\right)}+\dfrac{\left(c^2-1\right)\left(2c-\sqrt{c^2+3}\right)}{c\left(2c+\sqrt{c^2+3}\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(a^2-1\right)^2}{a\left(2a+\sqrt{a^2+3}\right)^2}+\dfrac{\left(b^2-1\right)^2}{b\left(2b+\sqrt{b^2+3}\right)^2}+\dfrac{\left(c^2-1\right)^2}{c\left(2c+\sqrt{c^2+3}\right)^2}\ge0\) (luôn đúng)
Khi \(f\left(t\right)=\sqrt{1+t}\) là hàm lõm trên \([-1, +\infty)\) ta có:
\(f(t)\le f(3)+f'(3)(t-3)\forall t\ge -1\)
Tức là \(f\left(t\right)\le2+\dfrac{1}{4}\left(t-3\right)=\dfrac{5}{4}+\dfrac{1}{4}t\forall t\ge-1\)
Áp dụng BĐT này ta có:
\(\sqrt{a^2+3}=a\sqrt{1+\dfrac{3}{a^2}}\le a\left(\dfrac{5}{4}+\dfrac{1}{4}\cdot\dfrac{3}{a^2}\right)=\dfrac{5}{4}a+\dfrac{3}{4}\cdot\dfrac{1}{a}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\sqrt{b^2+3}\le\dfrac{5}{4}b+\dfrac{3}{4}\cdot\dfrac{1}{b};\sqrt{c^2+3}\le\dfrac{5}{4}c+\dfrac{3}{4}\cdot\dfrac{1}{c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VP\le\dfrac{5}{4}\left(a+b+c\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=2\left(a+b+c\right)=VT\)
Đề bài hình như bị sai em, thay điểm rơi ko thỏa mãn
Biểu thức là \(a+b+\sqrt{2\left(a+c\right)}\) mới đúng
Bài 1:
Ta có:
\(\text{VT}=\frac{a^2}{a+2b^2}+\frac{b^2}{b+2c^2}+\frac{c^2}{c+2a^2}\)
\(=a-\frac{2ab^2}{a+2b^2}+b-\frac{2bc^2}{b+2c^2}+c-\frac{2ca^2}{c+2a^2}=(a+b+c)-2\left(\frac{ab^2}{a+2b^2}+\frac{bc^2}{b+2c^2}+\frac{ca^2}{c+2a^2}\right)\)
\(=3-2M(*)\)
Áp dụng BĐT Cauchy ta có:
\(M=\frac{ab^2}{a+b^2+b^2}+\frac{bc^2}{b+c^2+c^2}+\frac{ca^2}{c+a^2+a^2}\leq \frac{ab^2}{3\sqrt[3]{ab^4}}+\frac{bc^2}{3\sqrt[3]{bc^4}}+\frac{ca^2}{3\sqrt[3]{ca^4}}\)
\(\Leftrightarrow M\leq \frac{1}{3}(\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2})\)
Tiếp tục áp dụng BĐT Cauchy:
\(\sqrt[3]{a^2b^2}+\sqrt[3]{b^2c^2}+\sqrt[3]{c^2a^2}\leq \frac{ab+ab+1}{3}+\frac{bc+bc+1}{3}+\frac{ca+ca+1}{3}=\frac{2(ab+bc+ac)+3}{3}\)
Mà \(ab+bc+ac\leq \frac{(a+b+c)^2}{3}=3\) (quen thuộc)
\(\Rightarrow M\leq \frac{1}{3}.\frac{2.3+3}{3}=1(**)\)
Từ \((*);(**)\Rightarrow \text{VT}\geq 3-2.1=1\)
(đpcm)
Dấu bằng xảy ra khi $a=b=c=1$
Bài 2:
Áp dụng BĐT Cauchy -Schwarz:
\(\text{VT}=\frac{a^3}{a^2+a^2b^2}+\frac{b^3}{b^2+b^2c^2}+\frac{c^3}{c^2+a^2c^2}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{a^2+a^2b^2+b^2+b^2c^2+c^2+c^2a^2}\)
hay:
\(\text{VT}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{1+a^2b^2+b^2c^2+c^2a^2}(*)\)
Mặt khác, theo BĐT Cauchy ta dễ thấy:
\(a^4+b^4+c^4\geq a^2b^2+b^2c^2+c^2a^2\)
\(\Rightarrow (a^2+b^2+c^2)^2\geq 3(a^2b^2+b^2c^2+c^2a^2)\)
\(\Leftrightarrow 1\geq 3(a^2b^2+b^2c^2+c^2a^2)\Rightarrow a^2b^2+b^2c^2+c^2a^2\leq \frac{1}{3}(**)\)
Từ \((*);(**)\Rightarrow \text{VT}\geq \frac{(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2}{1+\frac{1}{3}}=\frac{3}{4}(a\sqrt{a}+b\sqrt{b}+c\sqrt{c})^2\)
Ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
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 \(a=b=c=1\)
\(\dfrac{a}{a+2\sqrt{\left(a+bc\right)}}=\dfrac{a}{a+2\sqrt{a\left(a+b+c\right)+bc}}=\dfrac{a}{a+2\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(=\dfrac{a}{a+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\)
\(\le\dfrac{a}{5^2}\left(\dfrac{1}{a}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\right)\)
\(=\dfrac{a}{25}\left(\dfrac{1}{a}+\dfrac{8}{\sqrt{\left(a+b\right)\left(a+c\right)}}\right)=\dfrac{1}{25}+\dfrac{8}{25}.\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
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 \(a=b=c=\dfrac{1}{3}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a^3}{\sqrt{b^2+3}}+\dfrac{a^3}{\sqrt{b^2+3}}+\dfrac{b^2+3}{8}\ge\dfrac{3a^2}{2}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b^3}{\sqrt{c^2+3}}+\dfrac{b^3}{\sqrt{c^2+3}}+\dfrac{c^2+3}{8}\ge\dfrac{3b^2}{2};\dfrac{c^3}{\sqrt{a^2+3}}+\dfrac{c^3}{\sqrt{a^2+3}}+\dfrac{a^2+3}{8}\ge\dfrac{3c^2}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(2P+\dfrac{a^2+b^2+c^2+9}{8}\ge\dfrac{3\left(a^2+b^2+c^2\right)}{2}\)
\(\Leftrightarrow P\ge\dfrac{\dfrac{3\left(a^2+b^2+c^2\right)}{2}-\dfrac{a^2+b^2+c^2+9}{8}}{2}=\dfrac{3}{2}\)
@DƯƠNG PHAN KHÁNH DƯƠNG
\(a;b;c\ge0\)thỏa mãn \(ab+bc+ca=1\). CMR \(\dfrac{1}{2a+2bc+1}+\dfrac{1}{2b+2ca+1}+\dfrac{1}{2c+2ab+1}\ge1\)
Đảm bảo an ninh :))