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\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)
Từ ( 1 ) và ( 2 ) có đpcm
a^4 +b^4 >= ab^3 +a^3 b (1)
<=> 4a^4 +4b^4 - 4ab(a^2 +b^2) >= 0
<=> [(a^2 +b^2 )^2 - 4ab(a^2 +a^2) +4a^2 b^2 ] +3a^4 +3b^4 -6a^2 b^2 >=0
<=> (a -b )^4 +3(a^4 + b^4 -2a^2 b^2 ) >= 0 (2)
cos (a-b )^4 >= 0
a^4 + b^4 >= 2a^2 b^2 (co si có thể không cần co si cũng được )
=> (2) đúng => (1) đúng => dpcm
b) a^2 +b^2 +1 >= ab +a+b (1)
<=>2a^2 +2b^2 +2 -2ab -2a-2b >=0
<=>[a^2 +b^2 -2ab ] +[a^2 -2a +1] +[b^2 -2b +1 ] >=0
<=>(a -b)^2 +(a-1)^2 + (b-1)^2 >=0 (2)
(2) đúng (1) đúng => dpcm
+) Ta có \(\sqrt{4a\left(3a+b\right)}\le\frac{4a+\left(3a+b\right)}{2}=\frac{7a+b}{2}\)
\(\Rightarrow\sqrt{a\left(3a+b\right)}\le\frac{7a+b}{4}\left(2\right)\)
+) Tương tự ta lại có :
\(\sqrt{b\left(3b+a\right)}\le\frac{7b+a}{4}\left(3\right)\)
+) Từ (2) và (3) ta có :
\(VT\left(1\right)\ge\frac{a+b}{\frac{7a+b}{4}+\frac{7b+a}{4}}=\frac{1}{2}\left(đpcm\right)\)
Ta có: \(\frac{a+b}{\sqrt{a\left(3a+b\right)}+\sqrt{b\left(3b+a\right)}}\)
\(=\frac{2\left(a+b\right)}{\sqrt{4a\left(3a+b\right)}+\sqrt{4b\left(3b+a\right)}}\ge\frac{2\left(a+b\right)}{\frac{1}{2}\left(4a+3a+b\right)+\frac{1}{2}\left(4b+3b+a\right)}\) (Cauchy)
\(=\frac{2\left(a+b\right)}{4\left(a+b\right)}=\frac{1}{2}\)
Dấu "=" xảy ra khi: a = b
Ta có
\(VT=\frac{\frac{1}{a^2}}{\frac{3}{c}+\frac{1}{b}}+\frac{\frac{1}{b^2}}{\frac{3}{a}+\frac{1}{c}}+\frac{\frac{1}{c^2}}{\frac{3}{b}+\frac{1}{a}}\)
Áp dụng bất đẳng thức buniacoxki dạng phân thức:
=> \(VT\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{4}{a}+\frac{4}{b}+\frac{4}{c}}=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{4}=504\)
Dấu bằng xảy ra khi a=b=c=3/2016
Lời giải:
Áp dụng BĐT Cauchy:
\(2\sqrt{a(3a+b)}=\sqrt{4a(3a+b)}\leq \frac{4a+3a+b}{2}\)
Tương tự \(2\sqrt{b(3b+a)}\leq \frac{4b+3b+a}{2}\)
\(\Rightarrow 2(\sqrt{a(3a+b)}+\sqrt{b(3b+a)})\leq \frac{8a+8b}{2}=4(a+b)\)
\(\Rightarrow \sqrt{a(3a+b)}+\sqrt{b(3b+a)}\leq 2(a+b)\)
\(\Rightarrow \frac{a+b}{\sqrt{a(3a+b)}+\sqrt{b(3b+a)}}\geq \frac{a+b}{2(a+b)}=\frac{1}{2}\) (đpcm)
Dấu bằng xảy ra khi \(a=b>0\)
Xét \(\frac{a^3}{a^2+ab+b^2}-\frac{b^3}{a^2+ab+b^2}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=a-b\)
Tương tự, ta được: \(\frac{b^3}{b^2+bc+c^2}-\frac{c^3}{b^2+bc+c^2}=b-c\); \(\frac{c^3}{c^2+ca+a^2}-\frac{a^3}{c^2+ca+a^2}=c-a\)
Cộng theo vế của 3 đẳng thức trên, ta được: \(\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)\)\(-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)=0\)
\(\Rightarrow\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\)\(=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\)
Ta đi chứng minh BĐT phụ sau: \(a^2-ab+b^2\ge\frac{1}{3}\left(a^2+ab+b^2\right)\)(*)
Thật vậy: (*)\(\Leftrightarrow\frac{2}{3}\left(a-b\right)^2\ge0\)*đúng*
\(\Rightarrow2LHS=\Sigma_{cyc}\frac{a^3+b^3}{a^2+ab+b^2}=\Sigma_{cyc}\text{ }\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}\)\(\ge\Sigma_{cyc}\text{ }\frac{\frac{1}{3}\left(a+b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=\frac{1}{3}\text{}\Sigma_{cyc}\left[\left(a+b\right)\right]=\frac{2\left(a+b+c\right)}{3}\)
\(\Rightarrow LHS\ge\frac{a+b+c}{3}=RHS\)(Q.E.D)
Đẳng thức xảy ra khi a = b = c
P/S: Có thể dùng BĐT phụ ở câu 3a để chứng minhxD:
1) ta chứng minh được \(\Sigma\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}=\Sigma\frac{b^4}{\left(a+b\right)\left(a^2+b^2\right)}\)
\(VT=\frac{1}{2}\Sigma\frac{a^4+b^4}{\left(a+b\right)\left(a^2+b^2\right)}\ge\frac{1}{4}\Sigma\frac{a^2+b^2}{a+b}\ge\frac{1}{8}\Sigma\left(a+b\right)=\frac{a+b+c+d}{4}\)
bài 2 xem có ghi nhầm ko
Áp dụng BĐT AM-GM ta có:
\(2\sqrt{a\left(3a+b\right)}=\sqrt{4a\left(3a+b\right)}\le\frac{4a+3a+b}{2}=\frac{7a+b}{2}\)
\(2\sqrt{b\left(3b+a\right)}=\sqrt{4b\left(3b+a\right)}\le\frac{4b+3b+a}{2}=\frac{7b+a}{2}\)
Suy ra \(\sqrt{b\left(3b+a\right)}+\sqrt{a\left(3a+b\right)}\le\frac{8a+8b}{4}=2\left(a+b\right)\)
\(\Rightarrow\frac{a+b}{\sqrt{b\left(3b+a\right)}+\sqrt{a\left(3a+b\right)}}\ge\frac{a+b}{2\left(a+b\right)}=\frac{1}{2}\)
\(\frac{4\left(a+b\right)}{2\sqrt{4a\left(3a+b\right)}+2\sqrt{4b\left(3b+a\right)}}\ge\frac{4\left(a+b\right)}{4a+3a+b+4b+3b+a}=\frac{4\left(a+b\right)}{8\left(a+b\right)}=\frac{1}{2}\)
Dấu "=" xảy ra khi \(a=b\)
a) \(a^3+b^3-a^2b-ab^2\)
\(=a^2\left(a-b\right)-b^2\left(a-b\right)\)
\(=\left(a^2-b^2\right)\left(a-b\right)=\left(a-b\right)^2\left(a+b\right)\ge0\)
\("="\Leftrightarrow a=b\)
b) \(a^4+b^4-a^3b-ab^3=a^3\left(a-b\right)-b^3\left(a-b\right)=\left(a^3-b^3\right)\left(a-b\right)\)
\(=\left(a-b\right)^2\left(a^2+\frac{1}{4}ab+b^2+\frac{3}{4}ab\right)=\left(a-b\right)^2\left[\left(a+\frac{1}{2}b\right)^2+\frac{3}{4}ab\right]\ge0\)
\("="\Leftrightarrow a=b\)