Cho a,b,c là các số dương.
a) CMR: \(a^3+b^3\ge a^2b+ab^2\)
b) Giả sử abc=1. Tìm GTLN của biểu thức:
\(P=\dfrac{1}{a^3+b^3+1}+\dfrac{1}{b^3+c^3+1}+\dfrac{1}{c^3+a^3+1}\)
Cho 3 số dương a,b,c thỏa mãn abc = 1. Tìm GTLN của biểu thức
\(P=\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\)
\(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2=2\left(ab+b+1\right)\)
Tương tự ...
\(\Rightarrow P\le\dfrac{1}{2\left(ab+b+1\right)}+\dfrac{1}{2\left(bc+c+1\right)}+\dfrac{1}{2\left(ca+a+1\right)}\)
\(=\dfrac{1}{2}\left(\dfrac{c}{abc+bc+c}+\dfrac{1}{bc+c+1}+\dfrac{bc}{ca.bc+a.bc+bc}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{c}{1+bc+c}+\dfrac{1}{bc+c+1}+\dfrac{bc}{c+1+bc}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{c+1+bc}{1+bc+c}\right)=\dfrac{1}{2}\)
\(P_{max}=\dfrac{1}{2}\) khi \(a=b=c=1\)
Cho a,b,c là số dương. CMR:
1. \(\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)
2. \(a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}\le a^3+b^3+c^3\)
3. \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
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 $a=b=c$
Bài 3:
Áp dụng BĐT Cauchy-Schwarz:
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{(a+b+c)^2}{b+c+c+a+a+b}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}$
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c$
Cho các số thực a,b,c thỏa mãn a>1 , b>\(\dfrac{1}{2}\) , \(c>\dfrac{1}{3}\) và \(\dfrac{1}{a}+\dfrac{2}{2b+1}+\dfrac{3}{3c+2}\ge2\). Tìm GTLN của bt \(P=\left(a-1\right)\left(2b-1\right)\left(3c-1\right)\)
Bài 1: Cho a,b,c là những số dương thỏa mãn: a+b+c=3
CMR: \(\dfrac{a^2}{a+2b^3}+\dfrac{b^2}{b+2c^3}+\dfrac{c^2}{c+2a^3}\ge1\)
Bài 2: Cho a, b, c thỏa mãn: ab+bc+ca=3
CMR: \(\dfrac{a}{2b^3+1}+\dfrac{b}{2c^3+1}+\dfrac{c}{2a^3+1}\ge1\)
Bài 3: Cho a, b, c > 0. CMR: \(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+3b\)
Dấu = xảy ra khi a=b=2c
Cho a,b,c là 3 số thức dương thỏa mãn a + b + c = 1/a + 1/b + 1/c . CMR
2( a + b + c) \(\ge\) \(\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}\)
Giải:
Dễ thấy bđt cần cm tương đương với mỗi bđt trong dãy sau:
\(\left(2a-\sqrt{a^2+3}\right)+\left(2b-\sqrt{b^2+3}\right)+\left(2c-\sqrt{c^2+3}\right)\ge0\),
\(\dfrac{a^2-1}{2a+\sqrt{a^2+3}}+\dfrac{b^2-1}{2b+\sqrt{b^2+3}}+\dfrac{c^2-1}{2c+\sqrt{c^2+3}}\ge0\),
\(\dfrac{\dfrac{a^2-1}{a}}{2+\sqrt{1+\dfrac{3}{a^2}}}+\dfrac{\dfrac{b^2-1}{b}}{2+\sqrt{1+\dfrac{3}{b^2}}}+\dfrac{\dfrac{c^2-1}{c}}{2+\sqrt{1+\dfrac{3}{b^2}}}\ge0\)
Các bđt trên đầu mang tính đối xứng giữa các biến nên k mất tính tổng quát ta có thể giả sử \(a\ge b\ge c\)
=> \(\dfrac{a^2-1}{a}\ge\dfrac{b^2-1}{b}\ge\dfrac{c^2-1}{c}\)
và \(\dfrac{1}{2+\sqrt{1+\dfrac{3}{a^2}}}\ge\dfrac{1}{2+\sqrt{1+\dfrac{3}{b^2}}}\ge\dfrac{1}{2+\sqrt{1+\dfrac{3}{c^2}}}\)
Áp dụng bđt Chebyshev có:
\(\dfrac{\dfrac{a^2-1}{a}}{2+\sqrt{1+\dfrac{3}{a^2}}}+\dfrac{\dfrac{b^2-1}{b}}{2+\sqrt{1+\dfrac{3}{b^2}}}+\dfrac{\dfrac{c^2-1}{c}}{2+\sqrt{1+\dfrac{3}{c^2}}}\ge\dfrac{1}{3}\left(\sum\dfrac{a^2-1}{a}\right)\left(\sum\dfrac{1}{2+\sqrt{1+\dfrac{3}{a^2}}}\right)\)
Theo gia thiết lại có: \(\sum\dfrac{a^2-1}{a}=\left(a+b+c\right)-\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=0\)
nên ta có thể suy ra \(\dfrac{\dfrac{a^2-1}{a}}{2+\sqrt{1+\dfrac{3}{a^2}}}+\dfrac{\dfrac{b^2-1}{b}}{2+\sqrt{1+\dfrac{3}{b^2}}}+\dfrac{\dfrac{c^2-1}{c}}{2+\sqrt{1+\dfrac{3}{c^2}}}\ge0\)
Vì vậy bđt đã cho ban đầu cũng đúng.
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:
\(VP\le\dfrac{5}{4}\left(a+b+c\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=2\left(a+b+c\right)=VT\)
cho a,b,c>0 thỏa mãn abc=1.CMR\(\dfrac{a^3}{1+b}+\dfrac{b^3}{1+c}+\dfrac{c^3}{1+a}\ge\dfrac{3}{2}\)
\(\dfrac{a^3}{1+b}+\dfrac{1+b}{4}+\dfrac{1}{2}\ge3\sqrt[3]{\dfrac{a^3\left(1+b\right)}{8\left(a+b\right)}}=\dfrac{3a}{2}\)
\(\dfrac{b^3}{1+c}+\dfrac{1+c}{4}+\dfrac{1}{2}\ge\dfrac{3b}{2}\) ; \(\dfrac{c^3}{1+a}+\dfrac{1+a}{4}+\dfrac{1}{2}\ge\dfrac{3c}{2}\)
\(\Rightarrow VT+\dfrac{a+b+c}{4}+\dfrac{9}{4}\ge\dfrac{3}{2}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{5}{4}\left(a+b+c\right)-\dfrac{9}{4}\ge\dfrac{5}{4}.3\sqrt[3]{abc}-\dfrac{9}{4}=\dfrac{3}{2}\)
cho 3 số a, b, c thoả mãn 0 < a, b, c < 1.CMR
\(\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\ge\dfrac{3}{3+abc}\)
Cho các số thực dương a,b. Tìm GTLN của biểu thức: \(P=\left(a+b\right)\left(\dfrac{1}{a^3+b}+\dfrac{1}{b^3+a}\right)-\dfrac{1}{ab}\)
\(\left(a^3+b\right)\left(\dfrac{1}{a}+b\right)\ge\left(a+b\right)^2\Rightarrow\dfrac{1}{a^3+b}\le\dfrac{\dfrac{1}{a}+b}{\left(a+b\right)^2}=\dfrac{ab+1}{a\left(a+b\right)^2}\)
Tương tự: \(\dfrac{1}{b^3+a}\le\dfrac{ab+1}{b\left(a+b\right)^2}\)
\(\Rightarrow P\le\left(a+b\right)\left(\dfrac{ab+1}{a\left(a+b\right)^2}+\dfrac{ab+1}{b\left(a+b\right)^2}\right)-\dfrac{1}{ab}\)
\(P\le\dfrac{\left(ab+1\right)}{a+b}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{1}{ab}=\dfrac{ab+1}{ab}-\dfrac{1}{ab}=1\)
\(P_{max}=1\) khi \(a=b=1\)
Cho a,b,c là các số thực k âm thỏa mãn a+b+c=3.CMR
a/ \(\dfrac{a}{1+b^2}+\dfrac{b}{1+c^2}+\dfrac{c}{1+a^2}\ge\dfrac{3}{2}\)
b/ \(\dfrac{a^2}{a+2b^3}+\dfrac{b^2}{b+2c^3}+\dfrac{c^2}{c+2a^3}\ge1\)
a) BĐT cần cm tương đương ;
\(a-\dfrac{ab^2}{1+b^2}+b-\dfrac{bc^2}{1+c^2}+a-\dfrac{a^2c}{1+a^2}\ge\dfrac{3}{2}\)
\(\Leftrightarrow3-\left(\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\right)\ge\dfrac{3}{2}\)
\(\Leftrightarrow\left(\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\right)\le\dfrac{3}{2}\)
Áp dụng BĐT Cauchy
\(\Rightarrow\dfrac{ab^2}{1+b^2}\le\dfrac{ab^2}{2b}=\dfrac{ab}{2}\)
tương tự rồi cộng vế theo vế các BĐT lại
\(\Leftrightarrow\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\le\dfrac{ab+bc+ac}{2}\)
mặt khác \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=3\)
\(\Rightarrow\dfrac{ab^2}{1+b^2}+\dfrac{bc^2}{1+c^2}+\dfrac{ac^2}{1+c^2}\le\dfrac{3}{2}\)
ĐPCM