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Bài 2:
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Trước hết ta chứng minh \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\)\(\Rightarrow1\ge\dfrac{2\sqrt{a\left(b+c\right)}}{a+b+c}\)
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\). Ta lại có:
\(\sqrt{\dfrac{a}{b+c}}=\dfrac{\sqrt{a}}{\sqrt{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)
Thiết lập các BĐT 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 theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\ge2\)
Dấu "=" không xảy ra nên ta có ĐPCM
Lưu ý: lần sau đăng từng bài 1 thôi nhé !
1) Áp dụng liên tiếp bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) với a;b là 2 số dương ta có:
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{4}\)\(\le\dfrac{\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{16}\)
TT: \(\dfrac{1}{a+2b+c}\le\dfrac{\dfrac{2}{b}+\dfrac{1}{a}+\dfrac{1}{c}}{16}\)
\(\dfrac{1}{a+b+2c}\le\dfrac{\dfrac{2}{c}+\dfrac{1}{a}+\dfrac{1}{b}}{16}\)
Cộng vế với vế ta được:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
Đặt \(\left\{{}\begin{matrix}x=a+b+c\\y=ab+bc+ca\end{matrix}\right.\) khi đó \(BDT\Leftrightarrow\dfrac{x^2+4x+y+3}{x^2+2x+y+xy}\le\dfrac{12+4x+y}{9+4x+2y}\)
\(\Leftrightarrow\dfrac{x^2+4x+y+3}{x^2+2x+y+xy}-1\le\dfrac{12+4x+y}{9+4x+2y}-1\)
\(\Leftrightarrow\dfrac{2x+3-xy}{x^2+2x+y+xy}\le\dfrac{3-y}{9+4x+2y}\)
\(\Leftrightarrow\dfrac{5x^2-3x^2y-xy^2-6xy+24x+y^2+3y+27}{\left(4x+2y+9\right)\left(x^2+xy+2x+y\right)}\le0\)
Đúng vì \(\dfrac{5}{3}x^2y\ge5x^2;\dfrac{x^2y}{3}\ge y^2;xy^2\ge9x;5xy\ge15x;xy\ge3y;x^2y\ge27\)
Áp dụng bất đẳng thức AM-GM ta có :
\(ab\le\frac{\left(a+b\right)^2}{4}\)=> \(\frac{ab}{a+b}\le\frac{a+b}{4}\)
Tương tự : \(\frac{bc}{b+c}\le\frac{b+c}{4}\); \(\frac{ca}{c+a}\le\frac{c+a}{4}\)
Cộng vế với vế các bđt trên ta có đpcm
Đẳng thức xảy ra <=> a=b=c
Có ab\le\dfrac{\left(a+b\right)^2}{4}\Rightarrow\dfrac{ab}{a+b}\le\dfrac{a+b}{4}ab≤4(a+b)2⇒a+bab≤4a+b.
Lời giải:
\(a+b+c=abc\Rightarrow a(a+b+c)=a^2bc\)
\(\Rightarrow a(a+b+c)+bc=bc(a^2+1)\)
\(\Leftrightarrow (a+b)(a+c)=bc(a^2+1)\)
\(\Leftrightarrow a^2+1=\frac{(a+b)(a+c)}{bc}\Rightarrow \frac{1}{\sqrt{a^2+1}}=\sqrt{\frac{bc}{(a+b)(a+c)}}\)
Áp dụng BĐT AM-GM:
\(\frac{1}{\sqrt{a^2+1}}=\sqrt{\frac{bc}{(a+b)(a+c)}}\leq \frac{1}{2}(\frac{b}{a+b}+\frac{c}{a+c})\)
Hoàn toàn tương tự:
\(\frac{1}{\sqrt{b^2+1}}=\sqrt{\frac{ac}{(b+a)(b+c)}}\leq \frac{1}{2}(\frac{a}{b+a}+\frac{c}{b+c})\)
\(\frac{1}{\sqrt{c^2+1}}=\sqrt{\frac{ab}{(c+a)(c+b)}}\leq \frac{1}{2}(\frac{a}{c+a}+\frac{b}{b+c})\)
Cộng theo vế:
\(\Rightarrow \frac{1}{\sqrt{a^2+1}}+\frac{1}{\sqrt{b^2+1}}+\frac{1}{\sqrt{c^2+1}}\leq \frac{1}{2}(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a})=\frac{3}{2}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\sqrt{3}$
1)
Kẻ phân giác AD,BK vuông góc với AD
sin A/2=sinBAD
xét tam giác AKB vuông tại K,có:
sinBAD=BK/AB (1)
xét tam giác BKD vuông tại K,có
BK<=BD thay vào (1):
sinBAD<=BD/AB(2)
lại có:BD/CD=AB/AC
=>BD/(BD+CD)=AB/(AB+AC)
=>BD/BC=AB/(AB+AC)
=>BD=(AB*BC)/(AB+AC) thay vào (2)
sinBAD<=[(AB*BC)/(AB+AC)]/AB
= BC/(AB + AC)
=>ĐPCM
làm rõ \(\sum_{cyc}\frac{a}{a+b}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b}{2(a+b)}\)
\(=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{2\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{c^2a-c^2b}{2\prod\limits_{cyc}(a+b)}\)
\(=\sum_{cyc}\frac{a^2b-a^2c}{2\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{2\prod\limits_{cyc}(a+b)}\geq0\) (đúng)
ok thỏa thuận rồi tui làm nửa sau thui nhé :D
Đặt \(a^2=x;b^2=y;c^2=z\) thì ta có:
\(VT=\sqrt{\dfrac{x}{x+y}}+\sqrt{\dfrac{y}{y+z}}+\sqrt{\dfrac{z}{x+z}}\)
Lại có: \(\sqrt{\dfrac{x}{x+y}}=\sqrt{\dfrac{x}{\left(x+y\right)\left(x+z\right)}\cdot\sqrt{x+z}}\)
Tương tự cộng theo vế rồi áp dụng BĐT C-S ta có:
\(VT^2\le2\left(x+y+z\right)\left[\dfrac{x}{\left(x+y\right)\left(x+z\right)}+\dfrac{y}{\left(y+z\right)\left(y+x\right)}+\dfrac{z}{\left(z+x\right)\left(z+y\right)}\right]\)
\(\Leftrightarrow VT^2\le\dfrac{4\left(x+y+z\right)\left(xy+yz+xz\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)
Vì \(VP^2=\dfrac{9}{2}\) nên cần cm \(VT\le \frac{9}{2}\)
\(\Leftrightarrow9\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8\left(x+y+z\right)\left(xy+yz+xz\right)\)
Can you continue
Lời giải:
Vì $a+b+c=1$ nên:
\(\text{VT}=\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\right)\)
\(=\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right)+\frac{3}{4}\)
\(=\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\right)+\frac{3}{4}\)
\(=(\frac{ab}{a^2+b^2}+\frac{a^2+b^2}{4ab})+(\frac{bc}{b^2+c^2}+\frac{b^2+c^2}{4bc})+(\frac{ca}{c^2+a^2}+\frac{c^2+a^2}{4ac})+\frac{3}{4}\)
\(\geq 2\sqrt{\frac{1}{4}}+2\sqrt{\frac{1}{4}}+2\sqrt{\frac{1}{4}}+\frac{3}{4}=\frac{15}{4}\) (áp dụng BĐT AM-GM)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
\(\dfrac{a^3}{b^3}+\dfrac{a^3}{b^3}+1+\dfrac{b^3}{c^3}+\dfrac{b^3}{c^3}+1+\dfrac{c^3}{a^3}+\dfrac{c^3}{a^3}+1\ge3\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\)
\(\Leftrightarrow2\left(\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\right)\ge3\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)-3\)
\(\ge2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)+3-3=2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\)
\(\Leftrightarrow\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)
giả sử \(a>b>c>0\) thì ta có :
\(\dfrac{a^2}{b^2}\left(\dfrac{a}{b}-1\right)+\dfrac{b^2}{c^2}\left(\dfrac{b}{c}-1\right)+\dfrac{c^2}{a^2}\left(\dfrac{c}{a}-1\right)\ge2\dfrac{a}{b}+\dfrac{c^2}{a^2}\left(\dfrac{c}{a}-1\right)\)
\(=\dfrac{2a}{b}+\dfrac{c^3}{a^3}-\dfrac{c^2}{a^2}\ge0\)
làm tương tự cho trường hợp \(c>b>a>0\) ; \(b>a>c\) và \(b>c>a\)
\(\Rightarrow\left(đpcm\right)\)
áp dụng bdt côsi \(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{3}{b}\)
tuông tu \(\dfrac{b^2}{c^3}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{3}{c}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{3}{a}\)
suy ra vt +\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
suy ra dpcm
dau = xay ra khi a=b=c
A B C D H K a, Vẽ phân giác AD của góc BAC
Kẻ BH\(\perp\)AD tại H ; CK\(\perp AD\) tại K
Dễ thấy \(sin\widehat{A_1}=sin\widehat{A_2}=sin\dfrac{A}{2}=\dfrac{BH}{AB}=\dfrac{CK}{AC}=\dfrac{BH+CK}{AB+AC}\le\)\(\le\dfrac{BD+CD}{b+c}=\dfrac{a}{b+c}\)
b, Tượng tự \(sin\dfrac{B}{2}\le\dfrac{b}{a+c};sin\dfrac{C}{2}\le\dfrac{c}{a+b}\)
Mặt khác \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}=8abc\)
\(\Rightarrow sin\dfrac{A}{2}.sin\dfrac{B}{2}.sin\dfrac{C}{2}\le\dfrac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{1}{8}\)
Ta có \(\dfrac{a^2}{b^2}+1\ge2.\dfrac{a}{b}\)
Lập 2 BĐT tương tự rồi cộng theo vế, ta được:
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}+3\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\) (*)
Mà ta lại có \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge3\sqrt[3]{\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{a}}=3\)
\(\Leftrightarrow-3\ge-\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\) (**)
Cộng theo vế (*) và (**), ta được đpcm.
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)