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Vì \(a^2+b^2\ge2ab,b^2+1\ge2b\),ta có:
\(\frac{1}{a^2+2b^2+3}=\frac{1}{a^2+b^2+b^2+1+1}\le\frac{1}{2\left(ab+b+1\right)}\)
Tương tự:\(\frac{1}{b^2+2c^2+3}\le\frac{1}{2\left(bc+c+1\right)}\)và \(\frac{1}{b^2+2c^2+3}\le\frac{1}{2\left(bc+c+1\right)}\)
Khi đó\(A\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+a}\right)\)
\(\Leftrightarrow A\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{ab+b+1}+\frac{b}{ab+b+1}\right)=\frac{1}{2}\)
Dấu"="trg BĐT trên xảy ra khi \(a=b=c=1\)
Vậy \(Max_P=\frac{1}{2}\Leftrightarrow a=b=c=1\)
Chắc không được GP đâu !!
Áp dụng bđt cauchy , ta có :
+) \(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2\)
+) \(b^2+2c^2+3\ge2bc+2c+2\)
+) \(c^2+2a^2+3\ge2ac+2a+2\)
Khi đó , ta có :
\(VT\le\frac{1}{2ab+2b+2}+\frac{1}{2bc+2c+2}+\frac{1}{2ac+2a+2}\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ac+a+1}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{abc}{bc+c+1}+\frac{abc}{ac+a+1}\right)\)( vì abc= 1 )
\(=\frac{1}{2}=VP\)( đoạn này ban tự phân tích ra nha , mk lmaf hơi tắt )
Vậy .................
2a^2 +2b^2 -5ab = 0
2a^2 -4ab -ab +2b^2 = 0
2a(a-2b) -b(a-2b) = 0
(2a-b)(a-2b) = 0
Suy ra: 2a=b hoặc a=2b
Mà a>b>0 nên a=2b
Ta có: P = a+b/a-b = 2b+b/ 2b-b = 3b/b=3
Vậy P = 3
Chúc bạn học tốt.
Ta có: \(2a^2+2b^2=5ab\)
\(\Leftrightarrow2a^2+2b^2-5ab=0\)
\(\Leftrightarrow2a^2-4ab-ab+2b^2=0\)
\(\Leftrightarrow2a\left(a-2b\right)-b\left(a-2b\right)=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a-2b=0\\2a-b=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=2b\\2a=b\end{cases}}}\)
Mà a > b > 0 nên a = 2b
Thế vào, ta được: \(P=\frac{a+b}{a-b}=\frac{2b+b}{2b-b}=\frac{3b}{b}=3\)
Vậy P = 3
\(2b=3-2a\)
\(P=\frac{2}{a}+\frac{1}{3-2a}=\frac{m\left(3-2a\right)}{a}+\frac{na}{3-2a}+k=\frac{9m-12ma+4ma^2+na^2+3ka-2ka^2}{a\left(3-2a\right)}=\frac{\left(4m+n-2k\right)a^2-3\left(4m-k\right)a+9m}{a\left(3-2a\right)}\)
\(=\frac{6-4a+a}{a\left(3-2a\right)}=\frac{-3a+6}{a\left(3-2a\right)}\)
=> 4m + n -2k =0 ; 4m -k = 1 ; 9m = 6
=> m= 2/3 ; k = 5/3 ; n= 2/3
\(P=\frac{2\left(3-2a\right)}{3a}+\frac{2a}{3\left(3-2a\right)}+\frac{5}{3}\ge2\sqrt{\frac{2\left(3-2a\right)}{3a}.\frac{2a}{3\left(3-2a\right)}}+\frac{5}{3}=3\)
P min = 3 khi 3-2a =a => a =1 ; b = 1/2
By Titu's Lemma we easy have:
\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)
\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)
\(=\frac{17}{4}\)
Mk xin b2 nha!
\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)
\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)
\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)
\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)
Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)
\(P=\frac{\frac{1}{a^2}}{\frac{1}{b}+\frac{1}{c}}+\frac{\frac{1}{b^2}}{\frac{1}{a}+\frac{1}{c}}+\frac{\frac{1}{c^2}}{\frac{1}{a}+\frac{1}{b}}\)
Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\Rightarrow xyz=1\Rightarrow P=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(P\ge\frac{\left(x+y+z\right)^2}{y+z+x+z+x+y}=\frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z\Leftrightarrow a=b=c=1\)
Cần cách khác thì nhắn cái
Gấp không có nghĩa là cho lên CHH nha :D
\(a^3-2a^2+a^2b+2a+2b=4\Leftrightarrow a^2\left(a+b\right)+2\left(a+b\right)-2a^2-4=0\Leftrightarrow\left(a^2+2\right)\left(a+b\right)-2\left(a^2+2\right)=0\Leftrightarrow\left(a^2+2\right)\left(a+b-2\right)=0\)
Vì \(a^2+2>0\forall a\)
\(\Rightarrow a+b-2=0\Leftrightarrow a+b=2\)
\(P=\frac{1}{a}+\frac{1}{b}\)
Áp dụng BĐT Cauchy Schwarz dạng Engel:
\(\Rightarrow P=\frac{1}{a}+\frac{1}{b}\ge\frac{\left(1+1\right)^2}{a+b}=\frac{4}{a+b}=\frac{4}{2}=2\)
Vậy \(Min_P=2\Leftrightarrow a=b=1\)
\(\Leftrightarrow a^2\left(a+b-2\right)+2\left(a+b-2\right)=0\)
\(\Leftrightarrow\left(a^2+2\right)\left(a+b-2\right)=0\)
\(\Leftrightarrow a+b=2\)
\(P=\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}=2\)