\(\left(a^3+b^3+c^3\right)\left[\dfrac{3}{8}.\left(a+b\right)\left(b+c\right)\left(c+a\right)\right].\left[\dfrac{3}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]..\)

\(\le\dfrac{1}{9^9}\left[a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^9\)

\(=\dfrac{1}{9^9}\left(a+b+c\right)^{27}\)

\(\Leftrightarrow3^8.\left(a^3+b^3+c^3\right)\le\dfrac{1}{3^{18}}\left(a+b+c\right)^{27}\)

\(\Leftrightarrow\sqrt[27]{\dfrac{a^3+b^3+c^3}{3}}\le\dfrac{a+b+c}{3}\)

P/s: Eztogiveup

Ara ara, dats a nais sol :3

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 \(a=b=c\)

Lời giải:

Đặt \(\left\{\begin{matrix} a+b-c=x\\ b+c-a=y\\ c+a-b=z\end{matrix}\right.\Rightarrow \left\{\begin{matrix} b=\frac{x+y}{2}\\ c=\frac{y+z}{2}\\ a=\frac{x+z}{2}\end{matrix}\right.\) \((x,y,z>0\) do $a,b,c$ là ba cạnh tam giác ).

BĐT cần chứng minh tương đương với :

\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq \frac{4}{(x+y)^2}+\frac{4}{(y+z)^2}+\frac{4}{(z+x)^2}\)

Áp dụng BĐT Cauchy:

\(\frac{1}{x^2}+\frac{1}{y^2}\geq \frac{2}{xy}\)

\(\Rightarrow 2\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\geq \left(\frac{1}{x}+\frac{1}{y}\right)^2\)

Theo BĐT S.Vacso: \(\frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}\Rightarrow 2\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\geq \frac{16}{(x+y)^2}(*)\)

Hoàn toàn tương tự:

\(2\left(\frac{1}{y^2}+\frac{1}{z^2}\right)\geq \frac{16}{(y+z)^2}; 2\left(\frac{1}{z^2}+\frac{1}{x^2}\right)\geq \frac{16}{(z+x)^2}(**)\)

Cộng theo vế \((*); (**)\) và rút gọn suy ra:

\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq \frac{4}{(x+y)^2}+\frac{4}{(y+z)^2}+\frac{4}{(z+x)^2}\)

Ta có đpcm

Dấu bằng xảy ra khi $x=y=z$ hay $a=b=c$

Lời giải:

Áp dụng BĐT Cauchy ta có:

\(\frac{a^4}{b^3(c+a)}+\frac{c+a}{4a}+\frac{1}{2}\geq 3\sqrt[3]{\frac{a^3}{8b^3}}=\frac{3a}{2b}\)

\(\frac{b^4}{c^3(a+b)}+\frac{a+b}{4b}+\frac{1}{2}\geq 3\sqrt[3]{\frac{b^3}{8c^3}}=\frac{3b}{2c}\)

\(\frac{c^4}{a^3(b+c)}+\frac{b+c}{4c}+\frac{1}{2}\geq 3\sqrt[3]{\frac{c^3}{8a^3}}=\frac{3c}{2a}\)

Cộng theo vế và rút gọn:

\(\Rightarrow \frac{a^4}{b^3(c+a)}+\frac{b^4}{c^3(a+b)}+\frac{c^4}{a^3(b+c)}+\frac{1}{4}(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})+\frac{9}{4}\geq \frac{3}{2}(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})\)

\(\Rightarrow \frac{a^4}{b^3(c+a)}+\frac{b^4}{c^3(a+b)}+\frac{c^4}{a^3(b+c)}\geq \frac{5}{4}(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})-\frac{9}{4}\geq \frac{5}{4}.3\sqrt[3]{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}-\frac{9}{4}\)

hay \( \frac{a^4}{b^3(c+a)}+\frac{b^4}{c^3(a+b)}+\frac{c^4}{a^3(b+c)}\geq \frac{5}{4}.3-\frac{9}{4}=\frac{3}{2}\)

Ta có đpcm

Dấu bằng xảy ra khi \(a=b=c\)

Cách khác:

Áp dụng BĐT Cauchy-Schwarz:

\(\text{VT}=\frac{(\frac{a^2}{b})^2}{b(c+a)}+\frac{(\frac{b^2}{c})^2}{c(a+b)}+\frac{(\frac{c^2}{a})^2}{a(b+c)}\geq \frac{\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2}{b(c+a)+c(a+b)+a(b+c)}\)

Tiếp tục áp dụng BĐT Cauchy-Schwarz:

\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq \frac{(a+b+c)^2}{b+c+a}=a+b+c\)

\(\Rightarrow \left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2\geq (a+b+c)^2\)

Do đó: \(\text{VT}\geq \frac{(a+b+c)^2}{2(ab+bc+ac)}\)

Theo hệ quả quen thuộc của BĐT Cauchy: \((a+b+c)^2\geq 3(ab+bc+ac)\)

Suy ra: \(\text{VT}\geq \frac{3(ab+bc+ac)}{2(ab+bc+ac)}=\frac{3}{2}\) (đpcm)

\(P=\dfrac{1}{a+bc}+\dfrac{1}{b+ca}+\left(a+b\right)\left(4+5c\right)\)

\(\ge\dfrac{4}{a+b+c\left(a+b\right)}+\left(a+b\right)\left(4+4c+c\right)\) (áp dụng \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)) \(=\dfrac{4}{\left(a+b\right)\left(1+c\right)}+4\left(a+b\right)\left(1+c\right)+\left(a+b\right)c\)

\(\ge2\sqrt{\dfrac{4}{\left(a+b\right)\left(1+c\right)}.4\left(a+b\right)\left(1+c\right)}+\left(a+b\right).0\) (áp dụng bđt côsi) \(=8+0=8\)

Dấu "=" xảy ra khi và chỉ khi a=b\(=\dfrac{1}{2};c=0\)

tks

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\((x+y)(x+z)\geq (x+\sqrt{yz})^2\)

\(\Rightarrow \sqrt{(x+y)(y+z)(x+z)}.\frac{\sqrt{y+z}}{x}\geq \frac{(y+z)(x+\sqrt{yz})}{x}=y+z+\frac{\sqrt{yz}(y+z)}{x}\)

Hoàn toàn tương tự :

\(\sqrt{(x+y)(y+z)(x+z)}.\frac{\sqrt{x+z}}{y}\geq x+z+\frac{\sqrt{xz}(x+z)}{y}\)

\(\sqrt{(x+y)(y+z)(x+z)}.\frac{\sqrt{x+y}}{z}\geq x+y+\frac{\sqrt{xy}(x+y)}{z}\)

Cộng theo vế:

\(T\geq 2(x+y+z)+\underbrace{\frac{(x+y)\sqrt{xy}}{z}+\frac{(y+z)\sqrt{yz}}{x}+\frac{(z+x)\sqrt{zx}}{y}}_{M}\)

Ta có:

\(M=\frac{(\sqrt{2}-z)\sqrt{xy}}{z}+\frac{(\sqrt{2}-x)\sqrt{yz}}{x}+\frac{(\sqrt{2}-y)\sqrt{xz}}{y}\)

\(=\sqrt{2}\left(\frac{\sqrt{xy}}{z}+\frac{\sqrt{yz}}{x}+\frac{\sqrt{xz}}{y}\right)-(\sqrt{xy}+\sqrt{yz}+\sqrt{xz})\)

Áp dụng BĐT AM-GM:

\(\frac{\sqrt{xy}}{z}+\frac{\sqrt{yz}}{x}+\frac{\sqrt{xz}}{y}\geq 3\sqrt[3]{\frac{xyz}{xyz}}=3\)

\(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\leq \frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=x+y+z=\sqrt{2}\)

Do đó: \(M\geq 3\sqrt{2}-\sqrt{2}=2\sqrt{2}\)

\(\Rightarrow T\geq 2(x+y+z)+M\geq 2\sqrt{2}+2\sqrt{2}=4\sqrt{2}\)

Vậy \(T_{\min}=4\sqrt{2}\)

Lời giải:

Với điều kiện đã cho thì hiển nhiên mẫu dương.

Áp dụng BĐT Cauchy-Schwarz ta có:

\(M=\frac{a^2}{2a\sqrt{b}-3a}+\frac{b^2}{2b\sqrt{c}-3b}+\frac{c^2}{2c\sqrt{a}-3c}\)\(\geq \frac{(a+b+c)^2}{2(a\sqrt{b}+b\sqrt{c}+c\sqrt{a})-3(a+b+c)}\)

Áp dụng BĐT Bunhiacopxky kết hợp BĐT AM-GM:

\((a\sqrt{b}+b\sqrt{c}+c\sqrt{a})^2\leq (a+b+c)(ab+bc+ac)\)

\(\leq (a+b+c).\frac{(a+b+c)^2}{3}=\frac{(a+b+c)^3}{3}\)

\(\Rightarrow a\sqrt{b}+b\sqrt{c}+c\sqrt{a}\leq \sqrt{\frac{(a+b+c)^3}{3}}\)

\(\Rightarrow M\geq \frac{(a+b+c)^2}{2\sqrt{\frac{(a+b+c)^3}{3}}-3(a+b+c)}\)

Đặt \(\sqrt{\frac{a+b+c}{3}}=t(t>\frac{3}{2})\)\(\Rightarrow a+b+c=3t^2\)

Ta có:

\(P\geq\frac{9t^4}{6t^3-9t^2}=\frac{3t^2}{2t-3}\)

\(\Leftrightarrow P\geq \frac{\frac{3}{4}(2t-3)(2t+3)}{2t-3}+\frac{27}{4(2t-3)}\)

\(\Leftrightarrow P\geq \frac{3}{4}(2t+3)+\frac{27}{4(2t-3)}=\frac{3}{4}(2t-3)+\frac{27}{4(2t-3)}+\frac{9}{2}\)

Áp dụng BĐT AM-GM:

\(\frac{3}{4}(2t-3)+\frac{27}{4(2t-3)}\geq 2\sqrt{\frac{3}{4}.\frac{27}{4}}=\frac{9}{2}\)

\(\Rightarrow P\geq \frac{9}{2}+\frac{9}{2}=9\)

Vậy \(P_{\min}=9\)

Đặt \(\left\{{}\begin{matrix}\sqrt{a}=x\\\sqrt{b}=y\\\sqrt{c}=z\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{x^2}{2y-3}+\dfrac{y^2}{2z-3}+\dfrac{z^2}{2x-3}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)-9}\ge9\)

Vì \(\dfrac{t^2}{2t-9}-9=\dfrac{\left(t-9\right)^2}{2t-9}\ge0\) (với \(t=x+y+z\))

Lời giải:

Áp dụng BĐT AM-GM:

\(3=x^2+y^2+\frac{1}{xy}\geq 2xy+\frac{1}{xy}\)

Đặt \(xy=t\Rightarrow 3\geq 2t+\frac{1}{t}\)

\(\Leftrightarrow 3t\geq 2t^2+1\Leftrightarrow 2t^2-3t+1\leq 0\)

\(\Leftrightarrow (2t-1)(t-1)\leq 0\Rightarrow \frac{1}{2}\leq t\leq 1\)

Với \(t=xy\leq 1\) ta có bổ đề sau:

\(\frac{1}{x^2+1}+\frac{1}{y^2+1}\leq \frac{2}{xy+1}(*)\)

Việc chứng minh bổ đề trên rất đơn giản. Thực hiện biến đổi tương đương và rút gọn ta thu được:

\((*)\Leftrightarrow (xy-1)(x-y)^2\leq 0\) (luôn đúng do \(xy\leq 1\) )

Áp dụng bổ đề trên vào bài toán đã cho:

\(P=2\left(\frac{1}{x^2+1}+\frac{1}{y^2+1}\right)-\frac{3}{2xy+1}\leq \frac{4}{xy+1}-\frac{3}{2xy+1}\)

\(\Leftrightarrow P\leq \frac{4}{t+1}-\frac{3}{2t+1}\)

Ta sẽ chứng minh \(\frac{4}{t+1}-\frac{3}{2t+1}\leq \frac{7}{6}\)

\(\Leftrightarrow \frac{5t+1}{2t^2+3t+1}\leq \frac{7}{6}\)

\(\Leftrightarrow 30t+6\leq 14t^2+21t+7\)

\(\Leftrightarrow 14t^2-9t+1\geq 0\)

\(\Leftrightarrow (2t-1)(7t-1)\geq 0\)

BĐT trên luôn đúng do \(t\geq \frac{1}{2}\)

Như vậy: \(P\leq \frac{4}{t+1}-\frac{3}{2t+1}\leq \frac{7}{6}\)

Vậy \(P_{\max}=\frac{7}{6}\). Dấu bằng xảy ra khi \(x=y=\frac{1}{\sqrt{2}}\)

tks

Lời giải:
Ta có:

\(a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(c+a)\)

\(=27-3(3-a)(3-b)(3-c)\)

\(=27-3[27-9(a+b+c)+3(ab+bc+ac)-abc]\)

\(=27-3[3(ab+bc+ac)-abc]=27-9(ab+bc+ac)+3abc\)

Do đó:

\(A=a^3+b^3+c^3+\frac{15}{4}abc=27-9(ab+bc+ac)+\frac{27}{4}abc(*)\)

Áp dụng BĐT Schur :

\(abc\geq (a+b-c)(b+c-a)(c+a-b)\)

\(\Leftrightarrow abc\geq (3-2a)(3-2b)(3-2c)\)

\(\Leftrightarrow abc\geq 27-18(a+b+c)+12(ab+bc+ac)-8abc\)

\(\Leftrightarrow 9abc\geq 12(ab+bc+ac)-27\)

\(\Leftrightarrow 3abc\geq 4(ab+bc+ac)-9\)

\(\Rightarrow \frac{27}{4}abc\geq 9(ab+bc+ac)-\frac{81}{4}(**)\)

Từ \((*); (**)\Rightarrow A\geq 27-\frac{81}{4}=\frac{27}{4}\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c=1\)

Ta có:

a³+b³+

==" nghiện zero 9 ak :))

Nghe mấy tiền bối đồn là đề này nằm trong đề đại học năm nào đó. Tự tìm nhá