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3.
\(\dfrac{2a^2}{b^2}+2\dfrac{b^2}{c^2}+2\dfrac{c^2}{a^2}\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
áp dụng bất đẳng thức cosi
+ \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\dfrac{a}{c}\)
......
tương tự với 2 cái sau
câu 1: \(VT=\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Bài 1:
Ta có: \(\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}=\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\)
Áp dụng bđt Cauchy Schwarz có:
\(\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{a\sqrt{a^2+8bc}+b\sqrt{b^2+8bc}+c\sqrt{c^2+8bc}}\)
Lại sử dụng bđt Cauchy schwarz ta có:
\(a\sqrt{a^2+8bc}+b\sqrt{b^2+8ac}+c\sqrt{c^2+8ab}=\sqrt{a}\cdot\sqrt{a^3+8abc}+\sqrt{b}\cdot\sqrt{b^3+8abc}+\sqrt{c}\cdot\sqrt{c^3+8abc}\ge\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}}=\sqrt{\dfrac{\left(a+b+c\right)^3}{a^3+b^3+c^3+24abc}}\)
=> Ta cần chứng minh: \(\left(a+b+c\right)^3\ge a^3+b^3+c^3+24abc\)
hay \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
Áp dụng bđt Cosi ta có:
\(a+b\ge2\sqrt{ab};b+c\ge2\sqrt{bc};c+a\ge2\sqrt{ca}\)
Nhân các vế của 3 bđt trên ta đc:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}=8\sqrt{a^2b^2c^2}=8abc\)
=> Đpcm
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a^3}{b\left(c+1\right)}+\dfrac{c+1}{4}+\dfrac{b}{2}\ge3\sqrt[3]{\dfrac{a^3}{b\left(c+1\right)}\cdot\dfrac{c+1}{4}\cdot\dfrac{b}{2}}\)
\(=3\sqrt[3]{\dfrac{a^3}{4\cdot2}\cdot\dfrac{c+1}{c+1}\cdot\dfrac{b}{b}}=3\sqrt[3]{\dfrac{a^3}{8}}=\dfrac{3a}{2}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b^3}{c\left(a+1\right)}\ge\dfrac{3b}{2};\dfrac{c^3}{a\left(b+1\right)}\ge\dfrac{3c}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT+\dfrac{a+b+c+3}{4}+\dfrac{a+b+c}{2}\ge\dfrac{3a+3b+3c}{2}\)
\(\Leftrightarrow VT+\dfrac{3\left(a+b+c\right)}{4}+\dfrac{3}{4}\ge\dfrac{3\left(a+b+c\right)}{2}\)
\(\Leftrightarrow VT+\dfrac{3}{4}\ge\dfrac{3\left(a+b+c\right)}{4}\). Mà theo AM-GM ta có:
\(a+b+c\ge3\sqrt[3]{abc}=3\)\(\Rightarrow VT+\dfrac{3}{4}\ge\dfrac{9}{4}\Rightarrow VT\ge\dfrac{3}{2}=VP\)
Đẳng thức xảy ra khi \(a=b=c=1\)
Bài 1
\(VT=\dfrac{a^2}{ab^2+abc+ac^2}+\dfrac{b^2}{c^2b+abc+a^2b}+\dfrac{c^2}{a^2c+abc+b^2c}\)
Áp dụng bđt Cauchy dạng phân thức
\(\Rightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{ab\left(a+b\right)+abc+ac\left(a+c\right)+abc+bc\left(b+c\right)+abc}\)
\(\Leftrightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{ab\left(a+b+c\right)+ac\left(a+b+c\right)+bc\left(a+b+c\right)}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}\)
\(\Leftrightarrow VT\ge\dfrac{a+b+c}{ab+bc+ac}\left(đpcm\right)\)
Dấu ''='' xảy ra khi \(a=b=c\)
Bài 2
\(VT=\left(\sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2}\right)\left[\left(\dfrac{\sqrt{a}}{b+c}\right)^2+\left(\dfrac{\sqrt{b}}{c+a}\right)^2+\left(\dfrac{\sqrt{c}}{a+b}\right)^2\right]\)
Áp dụng bđt Bunhiacopxki ta có
\(VT\ge\left(\sqrt{a}.\dfrac{\sqrt{a}}{b+c}+\sqrt{b}.\dfrac{\sqrt{b}}{c+a}+\sqrt{c}.\dfrac{\sqrt{c}}{a+b}\right)^2\)
\(\Leftrightarrow VT\ge\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\)
Xét \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
Áp dụng bđt Cauchy dạng phân thức ta có
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}=\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ac\right)}=\dfrac{3}{2}\)
\(\Rightarrow\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\ge\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\)
\(\Rightarrow VT\ge\dfrac{9}{4}\left(đpcm\right)\)
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:
\(VP\le\dfrac{5}{4}\left(a+b+c\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=2\left(a+b+c\right)=VT\)
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Bài 2:
Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:
\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)
\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)
\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
Bài 3:
Áp dụng BĐT AM-GM ta có:
\(\frac{a^2}{b^2}+1\geq 2\sqrt{\frac{a^2}{b^2}.1}=\frac{2a}{b}\). Tương tự:
\(\frac{b^2}{c^2}+1\geq \frac{2b}{c}; \frac{c^2}{a^2}+1\geq \frac{2c}{a}\)
Cộng các BĐT vừa thu được ta có :
\(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+3\geq 2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)
Mà theo BĐT AM-GM : \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3\sqrt[3]{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}=3\)
Suy ra: \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+3\geq 2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\)
\(\Rightarrow \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
------------------
Hoặc bạn có thể áp dụng BĐT Bunhiacopxky kết hợp AM-GM:
\(\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)(1+1+1)\geq \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\geq 3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)
\(\Rightarrow \left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\) (đpcm)
Bài 4:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\right)\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq \left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)^2\)
Mà theo kết quả bài 3 thì ta có:
\(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\). Do đó:
\(\left(\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\right)\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq \left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)
\(\Rightarrow \frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\geq \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c$
Bài 5:
Áp dụng BĐT AM-GM:
\((a+b+c)^2.\text{VT}=\frac{(a+b+c)^2}{a(a+b)}+\frac{(a+b+c)^2}{b(b+c)}+\frac{(a+b+c)^2}{c(c+a)}\)
\(=\frac{(a+b)^2+c^2+2c(a+b)}{a(a+b)}+\frac{a^2+(b+c)^2+2a(b+c)}{b(b+c)}+\frac{b^2+(c+a)^2+2b(c+a)}{c(c+a)}\)
\(=\frac{\frac{1}{4}(a+b)^2+c^2+2c(a+b)+\frac{3}{4}(a+b)^2}{a(a+b)}+\frac{a^2+\frac{1}{4}(b+c)^2+2a(b+c)+\frac{3}{4}(b+c)^2}{b(b+c)}+\frac{b^2+\frac{1}{4}(c+a)^2+2b(c+a)+\frac{3}{4}(c+a)^2}{c(c+a)}\)
\(\geq \frac{c(a+b)+2c(a+b)+\frac{3}{4}(a+b)^2}{a(a+b)}+\frac{a(b+c)+2a(b+c)+\frac{3}{4}(b+c)^2}{b(b+c)}+\frac{b(c+a)+2b(c+a)+\frac{3}{4}(c+a)^2}{c(c+a)}\)
\(=\frac{3c}{a}+\frac{3(a+b)}{4c}+\frac{3a}{b}+\frac{3(b+c)}{4b}+\frac{3b}{c}+\frac{3(c+a)}{4c}\)
\(=3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)+\frac{3}{4}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{c}+\frac{c}{b}+\frac{b}{a}\right)\)
\(\geq 3.3\sqrt[3]{1}+\frac{3}{4}.6\sqrt[6]{1}=\frac{27}{2}\)
Suy ra : \(\text{VT}\geq \frac{27}{2(a+b+c)^2}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Cách khác bài 5:
Áp dụng BĐT AM-GM ta có:
\(\text{VT}\geq 3\sqrt[3]{\frac{1}{a(a+b)}.\frac{1}{b(b+c)}.\frac{1}{c(c+a)}}=3\sqrt[3]{\frac{1}{a(b+c)b(c+a)c(a+b)}}=\frac{3}{\sqrt[3]{(ab+ac)(bc+ba)(ca+cb)}}(*)\)
Mà cũng theo BĐT AM-GM:
\((ab+ac)(bc+ba)(ca+cb)\leq \left(\frac{ab+ac+bc+ba+ca+cb}{3}\right)^3=\frac{8}{27}(ab+bc+ac)^3\)
\(ab+bc+ac\leq \frac{(a+b+c)^2}{3}\)
\(\Rightarrow (ab+ac)(bc+ba)(ca+cb)\leq \frac{8}{27}.\frac{(a+b+c)^6}{27}(**)\)
Từ \((*); (**)\Rightarrow \text{VT}\geq \frac{3}{\sqrt[3]{\frac{8}{27}.\frac{(a+b+c)^6}{27}}}=\frac{27}{2(a+b+c)^2}\)
Ta có đpcm.
Chỗ cộng 3 vế bạn nhầm rồi. \(2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\geq 2\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)\) mới đúng. Tuy nhiên đây lại không phải đpcm của bài toán.