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3: =>a^3+b^3+c^3>=3abc
=>(a+b)^3+c^3-3ab(a+b)-3abc>=0
=>(a+b+c)(a^2+b^2+c^2-ab-bc-ac)>=0
=>a^2+b^2+c^2-ab-bc-ac>=0
=>2a^2+2b^2+2c^2-2ab-2bc-2ac>=0
=>(a-b)^2+(a-c)^2+(b-c)^2>=0(luôn đúng)
Bài 3:
Ta có: \(a^2+b^2+c^2=3\ge ab+bc+ca\) ( tự cm bđt nha )
Áp dụng bất đẳng thức Schwarz ta có:
\(\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}=\dfrac{a^4}{ab+bc}+\dfrac{b^4}{bc+ab}+\dfrac{c^4}{ac+bc}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Dấu " = " khi a = b = c = 1
Bài 4:
Ta có: \(\dfrac{a^3}{a^2+b^2}=\dfrac{a\left(a^2+b^2\right)-ab^2}{a^2+b^2}=a-\dfrac{ab^2}{a^2+b^2}\ge a-\dfrac{ab^2}{2ab}=a-\dfrac{b}{2}\)
( BĐT AM - GM )
Tương tự \(\Rightarrow\dfrac{b^3}{c^2+a^2}\ge b-\dfrac{c}{2}\)
\(\dfrac{c^3}{c^2+a^2}\ge c-\dfrac{a}{2}\)
\(\Rightarrow VT\ge\left(a+b+c\right)-\dfrac{1}{2}\left(a+b+c\right)=\dfrac{a+b+c}{2}\)
Dấu " = " khi a = b = c
Tiếp sức cho Tú đệ
Bài 1: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\ge\left(a+b\right)\left(2ab-ab\right)=ab\left(a+b\right)\)
\(\Rightarrow\dfrac{a^3+b^3}{ab}\ge\dfrac{ab\left(a+b\right)}{ab}=a+b\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\ge VP."="\Leftrightarrow a=b=c\)
Bài 2: Holder:
\(\left(\dfrac{a^4}{bc^2}+\dfrac{b^4}{ca^2}+\dfrac{c^4}{ab^2}\right)\left(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}\right)\left(c+a+b\right)\ge\left(a+b+c\right)^3\)
Cần chứng minh \(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}\ge a+b+c\)
AM-GM: \(\dfrac{bc}{a}+\dfrac{ca}{b}\ge2\sqrt{\dfrac{bc}{a}\cdot\dfrac{ca}{b}}=2c\)
Tương tự rồi cộng theo vế:
\("=" \Leftrightarrow a=b=c\)
Bài 1:
Áp dụng BĐT Holder:
\((a^7+b^7+c^7)(a+b+c)(a+b+c)\geq (a^3+b^3+c^3)^3\)
\(\Rightarrow P=a^7+b^7+c^7\geq \frac{(a^3+b^3+c^3)^3}{(a+b+c)^2}\) \((1)\)
Tiếp tục Holder:
\((a^3+b^3+c^3)(1+1+1)(1+1+1)\geq (a+b+c)^3\)
\(\Rightarrow (a+b+c)\leq \sqrt[3]{9(a^3+b^3+c^3)}\) \((2)\)
Từ \((1),(2)\Rightarrow P\geq \frac{\sqrt[3]{(a^3+b^3+c^3)^7}}{\sqrt[3]{81}}\) \((3)\)
Áp dụng BĐT AM-GM:
\((a^3+b^3+c^3)^2\geq 3(a^3b^3+b^3c^3+c^3a^3)\geq 3\)
\(\Rightarrow a^3+b^3+c^3\geq \sqrt{3}\) \((4)\)
Từ \((3),(4)\Rightarrow P\geq \sqrt[6]{\frac{1}{3}}\)
Vậy \(P_{\min}=\sqrt[6]{\frac{1}{3}}\Leftrightarrow a=b=c=\sqrt[6]{\frac{1}{3}}\)
Bài 2:
Áp dụng BĐT AM-GM:
\(a^3+\sqrt{\frac{1}{27}}+\sqrt{\frac{1}{27}}\geq 3\sqrt[3]{a^3.\sqrt{\frac{1}{27^2}}}=a\)
\(b^3+\sqrt{\frac{1}{27}}+\sqrt{\frac{1}{27}}\geq 3\sqrt[3]{b^3.\sqrt{\frac{1}{27^2}}}=b\)
\(c^3+\sqrt{\frac{1}{27}}+\sqrt{\frac{1}{27}}\geq 3\sqrt[3]{c^3.\sqrt{\frac{1}{27^2}}}=c\)
Cộng theo vế:
\(a^3+b^3+c^3+6\sqrt{\frac{1}{27}}\geq a+b+c\)
Áp dụng BĐT AM-GM:
\((a+b+c)^2\geq 3(ab+bc+ac)=3\Rightarrow a+b+c\geq \sqrt{3}\)
Do đó, \(a^3+b^3+c^3\geq \sqrt{3}-6\sqrt{\frac{1}{27}}=\sqrt{\frac{1}{3}}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=\sqrt{\frac{1}{3}}\)
1/ a/dung bđt Cauchy - Schwarz dạng phân thức: \(\frac{a^2}{b+3c}+\frac{b^2}{c+3a}+\frac{c^2}{a+3b}\ge\frac{\left(a+b+c\right)^2}{4\left(a+b+c\right)}=\frac{a+b+c}{4}=\frac{3}{4}\)
2/ a/dung bđt bunhiacopxki :
\(S^2=\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+c+a\right)=3\cdot2\left(a+b+c\right)=6\cdot6=36\)
=> \(S\le6\)
Ta có: \(\frac{5a^3-b^3}{ab+3a^2}=\frac{3a^3-b^3}{ab+3a^2}+\frac{2a^3}{ab+3a^2}\)
\(=a-\frac{a^2b+b^3}{ab+3a^2}+\frac{2a^3}{ab+3a^2}\)
= \(a-\frac{b\left(a^2+b^2\right)}{a\left(b+3a\right)}+\frac{2a^3}{a\left(b+3a\right)}\) (1)
Áp dụng BĐT AM - GM ( x2 + y2 \(\ge2xy\)) ta có:
(1) \(\le a-\frac{2ab^2}{a\left(b+3a\right)}+\frac{2a^2}{b+3a}\) = \(a-\frac{2b^2}{b+3a}+\frac{2a^2}{b+3a}\) (2)
Tương tự ta cũng có:
\(\frac{5b^3-c^3}{bc+3b^2}\le b-\frac{2c^2}{c+3b}+\frac{2b^2}{c+3b}\left(3\right)\)
\(\frac{5c^3-a^2}{ca+3c^2}\)\(\le c-\frac{2a^2}{a+3c}+\frac{2c^2}{a+3c}\)(4)
Từ (2), (3), (4) \(\Rightarrow\frac{5a^3-b^3}{ab+3a^2}+\frac{5b^3-c^3}{bc+3b^2}+\frac{5c^3-a^3}{ca+3c^2}\le a+b+c+\left(\frac{2a^2}{a+3c}-\frac{2a^2}{a+3c}\right)+\left(\frac{2b^2}{b+3c}-\frac{2b^2}{b+3c}\right)+\left(\frac{2c^2}{c+3a}-\frac{2c^2}{c+3a}\right)=a+b+c\le2018\)
Vậy \(\frac{5a^3-b^3}{ab+3a^2}+\frac{5b^3-c^3}{bc+3b^2}+\frac{5c^3-a^3}{ca+3c^2}\le2018\)
Áp dụng bđt AM - GM:
\(T=\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{\sqrt[3]{abc}}{a+b+c}=\left(\dfrac{1}{9}\dfrac{a+b+c}{\sqrt[3]{abc}}+\dfrac{\sqrt[3]{abc}}{a+b+c}\right)+\dfrac{8}{9}\dfrac{a+b+c}{\sqrt[3]{abc}}\ge2\sqrt{\dfrac{1}{9}}+\dfrac{8}{9}.3=\dfrac{2}{3}+\dfrac{8}{3}=\dfrac{10}{3}\).
Đẳng thức xảy ra khi a = b = c.
Vậy Min T = \(\dfrac{10}{3}\) khi a = b = c.
BĐT 1 sai ngay với \(a=\sqrt{0,1},b=\sqrt{0,2},c=\sqrt{2,7}\)
BĐT 2 tương đương với đi chứng minh \(a^4b^4+b^4c^4+c^4a^4\geq 3a^2b^2c^2\)
Áp dụng BĐT AM-GM: \(a^4b^4+b^4c^4\geq 2a^2b^4c^2\)
Tương tự \(b^4c^4+c^4a^4\geq 2b^2c^4a^2,a^4b^4+c^4a^4\geq 2a^4b^2c^2\)
Cộng theo vế và rút gọn:
\(\Rightarrow a^4b^4+b^4c^4+c^4a^4\geq a^2b^2c^2(a^2+b^2+c^2)=3a^2b^2c^2\)
Do đó ta có đpcm. Dấu $=$ xảy ra khi $a=b=c=1$
thì ra cái đầu sai nghĩ mãi ko ra, đại ca thông minh thật :v