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Với x,y,z >0, ta có:
-\(\frac{x}{y}+\frac{y}{x}\ge2\) (1)
-\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\) (2)
-\(x^2+y^2+z^2\ge xy+yz+zx\Leftrightarrow\frac{x^2+y^2+z^2}{xy+yz+zx}\ge1\)(3)
Xảy ra đẳng thức ở (1),(2),(3) \(\Leftrightarrow x=y=z\), ta có:
\(P=\frac{ab+bc+ca}{a^2+b^2+c^2}+\left(a+b+c\right)^2.\frac{a+b+c}{abc}\)
=\(\frac{ab+bc+ca}{a^2+b^2+c^2}+\left(a^2+b^2+c^2+2ab+2bc+2ca\right).\frac{\left(a+b+c\right)}{abc}\)
Áp dụng các bất đẳng thức (1),(2),(3). ta có:
\(P\)\(\ge\)\(\frac{ab+bc+ca}{a^2+b^2+c^2}+\left(a^2+b^2+c^2\right).\frac{9}{ab+bc+ca}+2.9\)
=\(\left(\frac{ab+bc+ca}{a^2+b^2+c^2}+\frac{a^2+b^2+c^2}{ab+bc+ca}\right)+8.\frac{a^2+b^2+c^2}{ab+bc+ca}+18\)\(\ge2+8+18=28\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a^2+b^2+c^2=ab+bc+ca\\ab=bc=ca\end{cases}\Leftrightarrow a=b=c}\)
đặt biểu thức là P
2P+1 =\(\frac{2ab+2bc+2ca}{a^2+b^2+c^2}+1+\frac{2\left(a+b+c\right)^3}{abc}=\)\(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2}+\frac{2\left(a+b+c\right)^3}{abc}\)
\(\left(a+b+c\right)^2\left(\frac{1}{a^2+b^2+c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\right).\)(1)
Áp dụng bdt \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)dấu '=' khi x=y=z
(1) \(\ge\left(a+b+c\right)^2\left(\frac{1}{a^2+b^2+c^2}+\frac{18}{ab+bc+ca}\right)\)(2)
\(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}\ge\frac{9}{a^2+b^2+c^2+2ab+2bc+2ca}=\frac{9}{\left(a+b+c\right)^2}.\)
=> (2)\(\ge\left(a+b+c\right)^2\left(\frac{9}{\left(a+b+c\right)^2}+\frac{16}{ab+bc+ca}\right)=\)\(9+\frac{16\left(a+b+c\right)^2}{ab+bc+ca}\ge9+16.3=57\)
vì a2+b2+c2\(\ge ab+bc+ca< =>a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge3\left(ab+bc+ca\right)\)
<=> (a+b+c)2\(\ge\)3(ab+bc+ca)
vậy 2P+1\(\ge57< =>P\ge28\)
dấu '=' khi a=b=c
Áp dụng BĐT Cauchy-Schwarz ta có:
\((ab+a+1)^2 \le (a+b+c) \left( a+ a^2b+ \frac 1c \right) = (a+b+c)(a+a^2b+ab)\)
\(\Rightarrow \dfrac{a}{(ab+a+1)^2} \ge \dfrac{a}{(a+b+c)(a+a^2b+ab)}= \dfrac{1}{(a+b+c)(1+ab+b)}\)
Thiết lập các BĐT tương tự rồi cộng theo vế ta có:
\(\sum \dfrac{a}{(ab+a+1)^2} \ge \dfrac{1}{a+b+c} \sum \dfrac{1}{ab+b+1}= \dfrac{1}{a+b+c}\)
c2: Áp dụng BĐT bunyakovsky:
\(\left(a+b+c\right)\left[\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\right]\ge\left(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ca+c+1}\right)^2\)
Xét \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=\dfrac{a}{ab+a+1}+\dfrac{ab}{1+ab+a}+\dfrac{c}{c\left(a+1+ab\right)}\)
\(=\dfrac{ab+a+1}{ab+a+1}=1\)
do đó \(\left(a+b+c\right).VT\ge1\Leftrightarrow VT\ge\dfrac{1}{a+b+c}\)
dấu = xảy ra khi a=b=c=1
tu gia thiet co dc ab+bc+ca=0.Dat ab=x,bc=y,ca=z. Can chung minh x^3+y^3+z^3=3xyz
Ta có \(\sqrt{bc\left(1+a^2\right)}=\sqrt{bc+a^2bc}=\sqrt{bc+a\left(a+b+c\right)}\)
\(=\sqrt{\left(a+b\right)\left(a+c\right)}\)
Đặt BT đề cho là P
\(\Leftrightarrow P=\sum\dfrac{a}{\sqrt{bc\left(1+a^2\right)}}=\sum\sqrt{\dfrac{a}{a+b}\cdot\dfrac{a}{a+c}}\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{b+a}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\cdot3=\dfrac{3}{2}\)
Dấu \("="\Leftrightarrow a=b=c=\sqrt{3}\)
3: \(\left\{{}\begin{matrix}a+b>=2\sqrt{ab}\\b+c>=2\sqrt{bc}\\a+c>=2\sqrt{ac}\end{matrix}\right.\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)>=8abc\)
1: =>(a+b)(a^2-ab+b^2)-ab(a+b)>=0
=>(a+b)(a^2-2ab+b^2)>=0
=>(a+b)(a-b)^2>=0(luôn đúng)
Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\Rightarrow xyz=1\)
\(P=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)
\(M=\dfrac{\left(ab\right)^2}{abc^2\left(a+b\right)}+\dfrac{\left(ac\right)^2}{acb^2\left(a+c\right)}+\dfrac{\left(bc\right)^2}{a^2bc\left(b+c\right)}\)
\(M\ge\dfrac{\left(ab+bc+ca\right)^2}{2abc\left(ab+bc+ca\right)}=\dfrac{ab+bc+ca}{2abc}=\dfrac{\left(a+b+c\right)\left(ab+bc+ca\right)}{6abc}\ge\dfrac{9abc}{6abc}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Em nhớ mình đã làm bài này rồi mà sao nó ko hiển thì nhỉ:) Lười gõ lại nên copy bên AoPS luôn!
Equality holds when a = b = c
Link gốc: Inequality 99 (câu trả lời của SBM)
SMB = tth đấy:)) Ko phải ai khác đâu:)